Arithmetic Progressions

In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

(i) The taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each additional km.

(ii) The amount of air present in a cylinder when a vacuum pump removes $\dfrac{1}{4}$ of the air remaining in the cylinder at a time.

(iii) The cost of digging a well after every metre of digging, when it costs ₹150 for the first metre and rises by ₹50 for each subsequent metre.

(iv) The amount of money in the account every year, when ₹10000 is deposited at compound interest at 8 % per annum.

Answer

(i) It can be observed that

Taxi fare for 1st km = 15

Taxi fare for first 2 km = 15 + 8 = 23

Taxi fare for first 3 km = 23 + 8 = 31

Taxi fare for first 4 km = 31 + 8 = 39

Clearly 15, 23, 31, 39 ...... forms an A.P. because every term is 8 more than the preceding term.

Hence, the numbers involved forms an A.P. with common difference = 8.

(ii) Let the initial volume of air in a cylinder be V litre. In each stroke, the vacuum pump removes $\dfrac{1}{4}$ of air remaining in the cylinder at a time. In other words, after every stroke, only $1 - \dfrac{1}{4} = \dfrac{3}{4}$ part of air will remain.

Therefore, volumes will be $V, \dfrac{3}{4}V, \dfrac{3}{4}V \times \dfrac{3}{4}, \dfrac{3}{4}V \times \dfrac{3}{4} \times \dfrac{3}{4},$ ...........

⇒ $V, \dfrac{3}{4}V, \Big(\dfrac{3}{4}\Big)^2V, \Big(\dfrac{3}{4}\Big)^3V, ...........$

⇒ $V, \dfrac{3}{4}V, \dfrac{9}{16}V, \dfrac{27}{64}V, .......$

Difference between 2nd and 1st term :

$\dfrac{3}{4}V - V = \dfrac{3 - 4}{4}V = -\dfrac{1}{4}V$

Difference between 3rd and 2nd term :

$\dfrac{9}{16}V - \dfrac{3}{4}V = \dfrac{9 - 12}{16}V = -\dfrac{3}{16}V$

Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them.

Hence, numbers involved do not form an A.P.

(iii) Cost of digging for first metre = ₹150

Cost of digging for first 2 metres = ₹150 + ₹50 = ₹200

Cost of digging for first 3 metres = ₹200 + ₹50 = ₹250

Cost of digging for first 4 metres = ₹250 + ₹50 = ₹300

Clearly, 150, 200, 250, 300 ... forms an A.P. because every term is 50 more than the preceding term.

Hence, the numbers involved forms an A.P. with common difference = 50.

(iv) We know that,

If ₹P is deposited at r% compound interest per annum for n years, money will be $P\Big(1 + \dfrac{r}{100}\Big)^n$ after n years.

After every year, money will be :

$10000\Big(1 + \dfrac{8}{100}\Big), 10000\Big(1 + \dfrac{8}{100}\Big)^2, 10000\Big(1 + \dfrac{8}{100}\Big)^3, ..........$

Clearly, adjacent terms of this series do not have the same difference between them.

Hence, numbers involved do not form an A.P.

Write first four terms of the AP, when the first term a and the common difference d are given as follows:

(i) a = 10, d = 10

(ii) a = -2, d = 0

(iii) a = 4, d = -3

(iv) a = -1, d = $\dfrac{1}{2}$

(v) a = -1.25, d = -0.25

Answer

(i) First four terms of the A.P. are given by :

⇒ a, a + d, a + 2d, a + 3d.

Substituting values we get :

⇒ 10, 10 + 10, 10 + 2 × 10, 10 + 3 × 10.

⇒ 10, 20, 30, 40.

Hence, first four terms of the A.P. are 10, 20, 30 and 40.

(ii) First four terms of the A.P. are given by :

⇒ a, a + d, a + 2d, a + 3d.

Substituting values we get :

⇒ -2, -2 + 0, -2 + 2 × 0, -2 + 3 × 0.

⇒ -2, -2, -2, -2.

Hence, first four terms of the A.P. are -2, -2, -2, -2.

(iii) First four terms of the A.P. are given by :

⇒ a, a + d, a + 2d, a + 3d.

Substituting values we get :

⇒ 4, 4 + (-3), 4 + 2 × (-3) + 4 + 3 × (-3).

⇒ 4, 4 - 3, 4 - 6, 4 - 9.

⇒ 4, 1, -2, -5.

Hence, first four terms of the A.P. are 4, 1, -2, -5.

(iv) First four terms of the A.P. are given by :

⇒ a, a + d, a + 2d, a + 3d.

Substituting values we get :

⇒ -1, $-1 + \dfrac{1}{2}, -1 + 2 \times \dfrac{1}{2}, -1 + 3 \times \dfrac{1}{2}$

⇒ -1, $\dfrac{-2 + 1}{2}, -1 + 1, -1 + \dfrac{3}{2}$

⇒ -1, $-\dfrac{1}{2}, 0, \dfrac{-2 + 3}{2}$

⇒ -1, $-\dfrac{1}{2}, 0, \dfrac{1}{2}$.

Hence, first four terms of the A.P. are $-1, -\dfrac{1}{2}, 0, \dfrac{1}{2}$.

(v) First four terms of the A.P. are given by :

⇒ a, a + d, a + 2d, a + 3d.

Substituting values we get :

⇒ -1.25, -1.25 + (-0.25) + -1.25 + 2 × -0.25, -1.25 + 3 × -0.25

⇒ -1.25, -1.25 - 0.25, -1.25 + (-0.5), -1.25 + (-0.75)

⇒ -1.25, -1.5, -1.75, -2

Hence, first four terms of the A.P. are -1.25, -1.5, -1.75, -2.

For the following APs, write the first term and the common difference:

(i) 3, 1, -1, -3, .......

(ii) -5, -1, 3, 7, .......

(iii) $\dfrac{1}{3}, \dfrac{5}{3}, \dfrac{9}{3}, \dfrac{13}{3}$, .......

(iv) 0.6, 1.7, 2.8, 3.9, ........

Answer

(i) In A.P.,

3, 1, -1, -3, ........

First term (a) = 3

Common difference (d) = 1 - 3 = -2.

Hence, first term of the give A.P. is 3 and common difference is -2.

(ii) In A.P.,

-5, -1, 3, 7, .......

First term (a) = -5

Common difference (d) = -1 - (-5) = -1 + 5 = 4.

Hence, first term of the given A.P. is -5 and common difference is 4.

(iii) In A.P.,

$\dfrac{1}{3}, \dfrac{5}{3}, \dfrac{9}{3}, \dfrac{13}{3}$, .......

First term (a) = $\dfrac{1}{3}$

Common difference (d) = $\dfrac{5}{3} - \dfrac{1}{3} = \dfrac{4}{3}$.

Hence, first term of the give A.P. is $\dfrac{1}{3}$ and common difference is $\dfrac{4}{3}$.

(iv) In A.P.,

0.6, 1.7, 2.8, 3.9, ...........

First term (a) = 0.6

Common difference (d) = 1.7 - 0.6 = 1.1

Hence, first term of the given A.P. is 0.6 and common difference is 1.1

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.

(i) 2, 4, 8, 16, ....

(ii) 2, $\dfrac{5}{2}, 3, \dfrac{7}{2}, ......$

(iii) -1.2, -3.2, -5.2, -7.2, ........

(iv) -10, -6, -2, 2, .........

(v) 3, 3 + $\sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}$, ..........

(vi) 0.2, 0.22, 0.222, 0.2222, ..........

(vii) 0, -4, -8, -12, ........

(viii) $-\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}, ...........$

(ix) 1, 3, 9, 27, ........

(x) a, 2a, 3a, 4a, ..........

(xi) a, a², a³, a⁴, ...........

(xii) $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32},$ ...........

(xiii) $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12},$ ...........

(xiv) 1², 3², 5², 7², ......

(xv) 1², 5², 7², 73, ......

Answer

(i) Given,

2, 4, 8, 16, ...

⇒ a₂ - a₁ = 4 - 2 = 2,

⇒ a₃ - a₂ = 8 - 4 = 4,

Since, a₃ - a₂ ≠ a₂ - a₁

Hence, the given list of numbers do not form an A.P.

(ii) Given,

2, $\dfrac{5}{2}, 3, \dfrac{7}{2}, ......$

⇒ a₂ - a₁ = $\dfrac{5}{2} - 2 = \dfrac{5 - 4}{2} = \dfrac{1}{2}$,

⇒ a₃ - a₂ = $3 - \dfrac{5}{2} = \dfrac{6 - 5}{2} = \dfrac{1}{2}$,

⇒ a₄ - a₃ = $\dfrac{7}{2} - 3 = \dfrac{7 - 6}{2} = \dfrac{1}{2}$.

i.e., a_{k + 1} - a_k is same every time.

So, the given list of numbers forms A.P. with common difference = $\dfrac{1}{2}$.

Next three terms are :

⇒ $\dfrac{7}{2} + \dfrac{1}{2} = \dfrac{8}{2}$ = 4,

⇒ 4 + $\dfrac{1}{2} = \dfrac{8 + 1}{2} = \dfrac{9}{2}$.

⇒ $\dfrac{9}{2} + \dfrac{1}{2} = \dfrac{10}{2}$ = 5.

Hence, for the given A.P. common difference = $\dfrac{1}{2}$ and next three terms are 4, $\dfrac{9}{2}$ and 5.

(iii) Given,

-1.2, -3.2, -5.2, -7.2, ....

⇒ a₂ - a₁ = -3.2 - (-1.2) = -3.2 + 1.2 = -2,

⇒ a₃ - a₂ = -5.2 - (-3.2) = -5.2 + 3.2 = -2,

⇒ a₄ - a₃ = -7.2 - (-5.2) = -7.2 + 5.2 = -2.

i.e., a_{k + 1} - a_k is same every time.

So, the given list of numbers forms A.P. with common difference = -2.

Next three terms are :

⇒ -7.2 + (-2) = -7.2 - 2 = -9.2,

⇒ -9.2 + (-2) = -9.2 - 2 = -11.2

⇒ -11.2 + (-2) = -11.2 - 2 = -13.2

Hence, for the given A.P. common difference = -2 and next three terms are -9.2, -11.2 and -13.2

(iv) Given,

-10, -6, -2, 2, .........

⇒ a₂ - a₁ = -6 - (-10) = -6 + 10 = 4,

⇒ a₃ - a₂ = -2 - (-6) = -2 + 6 = 4,

⇒ a₄ - a₃ = 2 - (-2) = 2 + 2 = 4.

i.e., a_{k + 1} - a_k is same every time.

So, the given list of numbers forms A.P. with common difference = 4.

Next three terms are :

⇒ 2 + 4 = 6,

⇒ 6 + 4 = 10,

⇒ 10 + 4 = 14.

Hence, for the given A.P. common difference = 4 and next three terms are 6, 10 and 14

(v) Given,

3, 3 + $\sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}$, ..........

⇒ a₂ - a₁ = $3 + \sqrt{2} - 3 = \sqrt{2}$,

⇒ a₃ - a₂ = $3 + 2\sqrt{2} - (3 + \sqrt{2}) = \sqrt{2}$,

⇒ a₄ - a₃ = $3 + 3\sqrt{2} - (3 + 2\sqrt{2}) = \sqrt{2}$.

i.e., a_{k + 1} - a_k is same every time.

So, the given list of numbers forms A.P. with common difference = $\sqrt{2}$.

Next three terms are :

⇒ $3 + 3\sqrt{2} + \sqrt{2} = 3 + 4\sqrt{2}$,

⇒ $3 + 4\sqrt{2} + \sqrt{2} = 3 + 5\sqrt{2}$,

⇒ $3 + 5\sqrt{2} + \sqrt{2} = 3 + 6\sqrt{2}$.

Hence, for the given A.P. common difference = $\sqrt{2}$ and next three terms are $3 + 4\sqrt{2}, 3 + 5\sqrt{2}, 3 + 6\sqrt{2}$.

(vi) Given,

0.2, 0.22, 0.222, 0.2222, ........

⇒ a₂ - a₁ = 0.22 - 0.2 = 0.02,

⇒ a₃ - a₂ = 0.222 - 0.22 = 0.002.

Since, a₃ - a₂ ≠ a₂ - a₁

Hence, the given list of numbers do not form an A.P.

(vii) Given,

0, -4, -8, -12, ........

⇒ a₂ - a₁ = -4 - 0 = -4,

⇒ a₃ - a₂ = -8 - (-4) = -8 + 4 = -4,

⇒ a₄ - a₃ = -12 - (-8) = -12 + 8 = -4.

i.e., a_{k + 1} - a_k is same every time.

So, the given list of numbers forms A.P. with common difference = -4.

Next three terms are :

⇒ -12 + (-4) = -12 - 4 = -16,

⇒ -16 + (-4) = -16 - 4 = -20,

⇒ -20 + (-4) = -20 - 4 = -24.

Hence, for the given A.P. common difference = -4 and next three terms are -16, -20 and -24.

(viii) Given,

$-\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}, ...........$

⇒ a₂ - a₁ = $-\dfrac{1}{2} - \Big(-\dfrac{1}{2}\Big)$ = 0,

⇒ a₃ - a₂ = $-\dfrac{1}{2} - \Big(-\dfrac{1}{2}\Big)$ = 0,

⇒ a₄ - a₃ = $-\dfrac{1}{2} - \Big(-\dfrac{1}{2}\Big)$ = 0,

i.e., a_{k + 1} - a_k is same every time.

So, the given list of numbers forms A.P. with common difference = 0.

Next three terms are :

$\Rightarrow -\dfrac{1}{2} + 0 = -\dfrac{1}{2}$,

$\Rightarrow -\dfrac{1}{2} + 0 = -\dfrac{1}{2}$,

$\Rightarrow -\dfrac{1}{2} + 0 = -\dfrac{1}{2}$.

Hence, for the given A.P. common difference = 0 and next three terms are $-\dfrac{1}{2}, -\dfrac{1}{2}, -\dfrac{1}{2}$.

(ix) Given,

1, 3, 9, 27, .........

⇒ a₂ - a₁ = 3 - 1 = 2,

⇒ a₃ - a₂ = 9 - 3 = 6.

Since, a₃ - a₂ ≠ a₂ - a₁

Hence, the given list of numbers do not form an A.P.

(x) Given,

a, 2a, 3a, 4a, .....

⇒ a₂ - a₁ = 2a - a = a,

⇒ a₃ - a₂ = 3a - 2a = a,

⇒ a₄ - a₃ = 4a - 3a = a.

i.e., a_{k + 1} - a_k is same every time.

So, the given list of numbers forms A.P. with common difference = a.

Next three terms are :

⇒ 4a + a = 5a,

⇒ 5a + a = 6a,

⇒ 6a + a = 7a.

Hence, for the given A.P. common difference = a and next three terms are 5a, 6a and 7a.

(xi) Given,

a, a², a³, a⁴, .........

⇒ a₂ - a₁ = a² - a = a(a - 1),

⇒ a₃ - a₂ = a³ - a² = a²(a - 1).

Since, a₃ - a₂ ≠ a₂ - a₁

Hence, the given list of numbers do not form an A.P.

(xii) Given,

⇒ $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32},$ ...........

⇒ $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}$, ...........

⇒ a₂ - a₁ = $2\sqrt{2} - \sqrt{2} = \sqrt{2}$,

⇒ a₃ - a₂ = $3\sqrt{2} - 2\sqrt{2} = \sqrt{2}$,

⇒ a₄ - a₃ = $4\sqrt{2} - 3\sqrt{2} = \sqrt{2}$.

i.e., a_{k + 1} - a_k is same every time.

So, the given list of numbers forms A.P. with common difference = $\sqrt{2}$.

Next three terms are :

⇒ $4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{25 \times 2} = \sqrt{50}$,

⇒ $5\sqrt{2} + \sqrt{2} = 6\sqrt{2} = \sqrt{36 \times 2} = \sqrt{72}$,

⇒ $6\sqrt{2} + \sqrt{2} = 7\sqrt{2} = \sqrt{49 \times 2} = \sqrt{98}$.

Hence, for the given A.P. common difference = $\sqrt{2}$ and next three terms are $\sqrt{50}, \sqrt{72}, \sqrt{98}$.

(xiii) Given,

$\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12},$ ...........

⇒ a₂ - a₁ = $\sqrt{6} - \sqrt{3}$,

⇒ a₃ - a₂ = $\sqrt{9} - \sqrt{6}$.

Since, a₃ - a₂ ≠ a₂ - a₁

Hence, the given list of numbers do not form an A.P.

(xiv) Given,

⇒ 1², 3², 5², 7², ......

⇒ 1, 9, 25, 49, .........

⇒ a₂ - a₁ = 9 - 1 = 8,

⇒ a₃ - a₂ = 25 - 9 = 16.

Since, a₃ - a₂ ≠ a₂ - a₁

Hence, the given list of numbers do not form an A.P.

(xv) Given,

⇒ 1², 5², 7², 73, ......

⇒ 1, 25, 49, 73, .........

⇒ a₂ - a₁ = 25 - 1 = 24,

⇒ a₃ - a₂ = 49 - 25 = 24,

⇒ a₄ - a₃ = 73 - 49 = 24.

i.e., a_{k + 1} - a_k is same every time.

So, the given list of numbers forms A.P. with common difference = 24.

Next three terms are :

⇒ 73 + 24 = 97,

⇒ 97 + 24 = 121,

⇒ 121 + 24 = 145.

Hence, for the given A.P. common difference = 24 and next three terms are 97, 121 and 145.

Fill in the blanks in the following table, given that a is the first term, d the common difference and a_n the nth term of the AP :

Answer

(i) By formula,

a_n = a + (n - 1)d

Substituting values we get :

a_n = 7 + 3(8 - 1)

= 7 + 3 × 7

= 7 + 21

= 28.

Hence, a_n = 28.

(ii) By formula,

a_n = a + (n - 1)d

Substituting values we get :

⇒ 0 = -18 + (10 - 1)d

⇒ 0 = -18 + 9d

⇒ 9d = 18

⇒ d = $\dfrac{18}{9}$ = 2.

Hence, d = 2.

(iii) By formula,

a_n = a + (n - 1)d

Substituting values we get :

⇒ -5 = a + (18 - 1)(-3)

⇒ -5 = a + 17 × -3

⇒ -5 = a - 51

⇒ a = 51 - 5 = 46.

Hence, a = 46.

(iv) By formula,

a_n = a + (n - 1)d

Substituting values we get :

⇒ 3.6 = -18.9 + (n - 1)(2.5)

⇒ 3.6 = -18.9 + 2.5n - 2.5

⇒ 3.6 = 2.5n - 21.4

⇒ 2.5n = 21.4 + 3.6

⇒ 2.5n = 25

⇒ n = $\dfrac{25}{2.5}$ = 10.

Hence, n = 10.

(v) By formula,

a_n = a + (n - 1)d

Substituting values we get :

a_n = 3.5 + 0(105 - 1)

= 3.5 + 0 × 104

= 3.5 + 0

= 3.5

Hence, a_n = 3.5

Choose the correct choice in the following and justify :

30th term of the A.P. : 10, 7, 4, ....., is

1. 97

2. 77

3. -77

4. -87

Answer

Given,

A.P. : 10, 7, 4, .....,

First term (a) = 10 and common difference (d) = 7 - 10 = -3.

By formula,

a_n = a + (n - 1)d

Substituting values we get :

a₃₀ = 10 + (30 - 1)(-3)

= 10 + 29 × -3

= 10 - 87

= -77.

Hence, Option 3 is the correct option.

Choose the correct choice in the following and justify :

11th term of the A.P. : -3, $-\dfrac{1}{2}$, 2, ....., is

1. 28

2. 22

3. -38

4. $-48\dfrac{1}{2}$

Answer

Given,

A.P. : -3, $-\dfrac{1}{2}$, 2, .....,

First term (a) = -3 and common difference (d) = $-\dfrac{1}{2} - (-3) = -\dfrac{1}{2} + 3 = 2\dfrac{1}{2}$ = 2.5

By formula,

a_n = a + (n - 1)d

Substituting values we get :

a₁₁ = -3 + (11 - 1) × 2.5

= -3 + 10 × 2.5

= -3 + 25

= 22.

Hence, Option 2 is the correct option.

In the following APs, find the missing terms :

(i) 2, ......, 26.

(ii) ......, 13, ......, 3.

(iii) 5, ......, ......., $9\dfrac{1}{2}$

(iv) -4, ......, ....., ....., ......, 6

(v) ......, 38, ......, ....., ......, -22

Answer

(i) Let x be the missing term.

A.P. : 2, x, 26

Since, above list is in A.P. so common difference will be equal among consecutive numbers.

∴ x - 2 = 26 - x

⇒ x + x = 26 + 2

⇒ 2x = 28

⇒ x = $\dfrac{28}{2}$ = 14.

Hence, missing term = 14.

(ii) Let x and y be the missing term.

A.P. : x, 13, y, 3.

Since, above list is in A.P. so common difference will be equal among consecutive numbers.

∴ 3 - y = y - 13

⇒ y + y = 13 + 3

⇒ 2y = 16

⇒ y = $\dfrac{16}{2}$

⇒ y = 8.

Common difference (d) : y - 13 = 8 - 13 = -5.

⇒ 13 - x = d

⇒ 13 - x = -5

⇒ x = 13 + 5 = 18

Hence, missing terms are 18 and 8.

(iii) Let x and y be the missing term.

A.P. : 5, x, y, $9\dfrac{1}{2}$

In above A.P.,

First term (a) = 5

Fourth term (a₄) = $9\dfrac{1}{2}$

By formula,

a_n = a + (n - 1)d

$\Rightarrow a_4 = 5 + (4 - 1) \times d \\[1em] \Rightarrow 9\dfrac{1}{2} = 5 + 3d \\[1em] \Rightarrow \dfrac{19}{2} - 5 = 3d \\[1em] \Rightarrow 3d = \dfrac{19- 10}{2} \\[1em] \Rightarrow 3d = \dfrac{9}{2} \\[1em] \Rightarrow d = \dfrac{9}{2} \times \dfrac{1}{3} \\[1em] \Rightarrow d = \dfrac{3}{2}.$

In above A.P.

x = 2nd term = a + d = $5 + \dfrac{3}{2} = \dfrac{10 + 3}{2} = \dfrac{13}{2} = 6\dfrac{1}{2}$.

y = 3rd term = a + 2d = $5 + 2 \times \dfrac{3}{2}$ = 5 + 3 = 8.

Hence, missing terms are $6\dfrac{1}{2}, 8$.

(iv) By formula,

a_n = a + (n - 1)d

First term (a) = -4

Sixth term (a₆) = 6

⇒ a + (6 - 1)d = 6

⇒ -4 + 5d = 6

⇒ 5d = 6 + 4

⇒ 5d = 10

⇒ d = $\dfrac{10}{5}$ = 2.

Second term = a + d = -4 + 2 = -2

Third term = a + 2d = -4 + 4 = 0

Fourth term = a + 3d = -4 + 6 = 2

Fifth term = a + 4d = -4 + 8 = 4

Hence, the missing terms are -2, 0, 2, 4.

(v) By formula,

a_n = a + (n - 1)d

Let the first term be a and common difference be d.

Given,

Second term = 38

⇒ a + (2 - 1)d = 38

⇒ a + d = 38 ....(1)

Given,

Sixth term = -22

⇒ a + (6 - 1)d = -22

⇒ a + 5d = -22 ....(2)

On Subtracting equation (1) from equation (2), we get :

⇒ a + 5d - (a + d) = -22 - 38

⇒ a - a + 5d - d = -60

⇒ 4d = -60

⇒ d = $\dfrac{-60}{4}$ = -15.

Substituting value of d in equation (1), we get :

⇒ a + (-15) = 38

⇒ a = 38 + 15 = 53.

Third term = a₃

= a + (3 - 1)d

= a + 2d

= 53 + 2(-15)

= 53 - 30 = 23.

Fourth term = a₄

= a + (4 - 1)d

= a + 3d

= 53 + 3(-15)

= 53 - 45

= 8.

Fifth term = a₅

= a + (5 - 1)d

= a + 4d

= 53 + 4(-15)

= 53 - 60

= -7.

Hence, the missing terms are 53, 23, 8, -7.

Which term of the AP : 3, 8, 13, 18, ......., is 78 ?

Answer

In above A.P.,

3, 8, 13, 18, .......

First term (a) = 3

Common difference (d) = 8 - 3 = 5.

Let nth term of the A.P. be 78.

By formula,

⇒ a_n = a + (n - 1)d

Substituting values we get :

⇒ 78 = 3 + (n - 1) × 5

⇒ 78 = 3 + 5n - 5

⇒ 78 = 5n - 2

⇒ 5n = 78 + 2

⇒ 5n = 80

⇒ n = $\dfrac{80}{5}$

⇒ n = 16.

Hence, 16th term of the A.P. is 78.

Find the number of terms in each of the following APs :

(i) 7, 13, 19, ......., 205

(ii) 18, $15\dfrac{1}{2}, 13, ......., -47$

Answer

(i) Given,

7, 13, 19, ......., 205.

In the above A.P.,

First term (a) = 7 and common difference (d) = 13 - 7 = 6.

Let nth term be 205.

By formula,

a_n = a + (n - 1)d

Substituting values we get :

⇒ 205 = 7 + 6(n - 1)

⇒ 205 = 7 + 6n - 6

⇒ 205 = 6n + 1

⇒ 6n = 205 - 1

⇒ 6n = 204

⇒ n = $\dfrac{204}{6}$ = 34.

Hence, number of terms = 34.

(ii) Given,

18, $15\dfrac{1}{2}, 13, ......., -47$

In the above A.P.,

First term (a) = 18,

Common difference (d) = $15\dfrac{1}{2} - 18 = \dfrac{31}{2} - 18 = \dfrac{31 - 36}{2} = -\dfrac{5}{2}$.

Let nth term be -47.

By formula,

a_n = a + (n - 1)d

Substituting values we get :

⇒ -47 = $18 + (n - 1) \times -\dfrac{5}{2}$

⇒ -47 = $18 + \dfrac{-5n + 5}{2}$

⇒ -47 = $\dfrac{36 - 5n + 5}{2}$

⇒ -94 = 41 - 5n

⇒ 5n = 41 + 94

⇒ 5n = 135

⇒ n = $\dfrac{135}{5}$ = 27.

Hence, no. of terms = 27.

Check whether -150 is a term of the AP : 11, 8, 5, 2 .......

Answer

In the A.P. : 11, 8, 5, 2 .......

First term (a) = 11 and common difference (d) = 8 - 11 = -3.

Let -150 be nth term of the A.P.

By formula,

a_n = a + (n - 1)d

Substituting values we get :

⇒ -150 = 11 + (n - 1) × -3

⇒ -150 = 11 - 3n + 3

⇒ 3n = 150 + 11 + 3

⇒ 3n = 164

⇒ n = $\dfrac{164}{3} = 54\dfrac{2}{3}$.

Since, number of term cannot be in fraction.

Hence, -150 cannot be the term of the A.P.

Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Answer

Let first term be a and common difference be d.

By formula,

a_n = a + (n - 1)d

Given,

⇒ a₁₁ = 38

⇒ a + (11 - 1)d = 38

⇒ a + 10d = 38 .......(1)

Given,

⇒ a₁₆ = 73

⇒ a + (16 - 1)d = 73

⇒ a + 15d = 73 ........(2)

Subtracting equation (1) from (2), we get :

⇒ a + 15d - (a + 10d) = 73 - 38

⇒ a - a + 15d - 10d = 35

⇒ 5d = 35

⇒ d = $\dfrac{35}{5}$

⇒ d = 7.

Substituting value of d in equation (1), we get :

⇒ a + 10 × 7 = 38

⇒ a + 70 = 38

⇒ a = 38 - 70 = -32.

31st term = a₃₁

= a + (31 - 1)d

= -32 + 30 × 7

= -32 + 210 = 178.

Hence, 31st term of A.P. = 178.

An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Answer

Let first term of A.P. be a and common difference be d.

By formula,

a_n = a + (n - 1)d

Given,

Last term of A.P. = 106 and there are 50 terms in the A.P.

⇒ a₅₀ = 106

⇒ a + (50 - 1)d = 106

⇒ a + 49d = 106 .......(1)

Given,

3rd term = 12

⇒ a₃ = 12

⇒ a + (3 - 1)d = 12

⇒ a + 2d = 12 .......(2)

Subtracting equation (2) from (1), we get :

⇒ a + 49d - (a + 2d) = 106 - 12

⇒ a - a + 49d - 2d = 94

⇒ 47d = 94

⇒ d = $\dfrac{94}{47}$ = 2.

Substituting value of d in equation (2), we get :

⇒ a + 2 × 2 = 12

⇒ a + 4 = 12

⇒ a = 12 - 4

⇒ a = 8.

29th term = a₂₉

= a + (29 - 1)d

= 8 + 28 × 2

= 8 + 56

= 64.

Hence, 29th term = 64.

If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?

Answer

Let first term of A.P. be a and common difference be d.

By formula,

a_n = a + (n - 1)d

Given,

3rd term = 4

⇒ a₃ = 4

⇒ a + (3 - 1)d = 4

⇒ a + 2d = 4 ........(1)

Given,

9th term = -8

⇒ a₉ = -8

⇒ a + (9 - 1)d = -8

⇒ a + 8d = -8 ........(2)

Subtracting equation (1) from (2), we get :

⇒ a + 8d - (a + 2d) = -8 - 4

⇒ a - a + 8d - 2d = -12

⇒ 6d = -12

⇒ d = $-\dfrac{12}{6}$ = -2.

Substituting value of d in equation (1), we get :

⇒ a + 2 × -2 = 4

⇒ a - 4 = 4

⇒ a = 4 + 4 = 8.

Let nth term of the A.P. be zero.

⇒ a_n = a + (n - 1)d

⇒ 0 = 8 + (n - 1) × (-2)

⇒ 0 = 8 - 2n + 2

⇒ 2n = 10

⇒ n = $\dfrac{10}{2}$ = 5.

Hence, 5th term of the A.P. is zero.

The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Answer

Let first term of A.P. be a and common difference be d.

By formula,

a_n = a + (n - 1)d

Given,

17th term of an AP exceeds its 10th term by 7.

∴ a₁₇ - a₁₀ = 7

⇒ a + (17 - 1)d - [a + (10 - 1)d] = 7

⇒ a + 16d - [a + 9d] = 7

⇒ a - a + 16d - 9d = 7

⇒ 7d = 7

⇒ d = $\dfrac{7}{7}$ = 1.

Hence, common difference = 1.

Which term of the AP : 3, 15, 27, 39,....... will be 132 more than its 54th term ?

Answer

In the given A.P.,

First term (a) = 3 and common difference (d) = 15 - 3 = 12.

By formula,

a_n = a + (n - 1)d.

Let nth term of A.P. be 132 more than 54th term.

∴ a_n - a₅₄ = 132

⇒ a + (n - 1)d - [a + (54 - 1)d] = 132

⇒ 3 + 12(n - 1) - [3 + 53 × 12] = 132

⇒ 3 + 12n - 12 - 3 - 636 = 132

⇒ 12n - 648 = 132

⇒ 12n = 132 + 648

⇒ 12n = 780

⇒ n = $\dfrac{780}{12}$ = 65.

Hence, required term = 65th.

Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Answer

Let first term of first A.P. be a₁

and

first term of second A.P. be b₁.

Let common difference be d.

For 1st A.P. :

100th term = a₁ + (100 - 1)d

= a₁ + 99d.

For 2nd A.P. :

100th term = b₁ + (100 - 1)d

= b₁ + 99d.

Given,

Difference between 100th terms is 100 :

∴ b₁ + 99d - (a₁ + 99d) = 100

⇒ b₁ - a₁ + 99d - 99d = 100

⇒ b₁ - a₁ = 100 ..........(1)

For 1st A.P. :

1000th term = a₁ + (1000 - 1)d

= a₁ + 999d.

For 2nd A.P. :

1000th term = b₁ + (1000 - 1)d

= b₁ + 999d.

Difference between 1000th terms :

⇒ b₁ + 999d - (a₁ + 999d)

⇒ b₁ - a₁ + 999d - 999d

⇒ b₁ - a₁

⇒ 100 [From (1)]

Hence, difference between 1000 terms of the A.P. will be 100.

How many three-digit numbers are divisible by 7?

Answer

List of the three digit numbers divisible by 7 are :

105, 112, 119, ........., 994.

The above list is an A.P. with,

First term (a) = 105 and common difference (d) = 112 - 105 = 7.

Let nth term be 994.

By formula,

⇒ a_n = a + (n - 1)d

⇒ 994 = 105 + 7(n - 1)

⇒ 994 = 105 + 7n - 7

⇒ 994 = 7n + 98

⇒ 7n = 994 - 98

⇒ 7n = 896

⇒ n = $\dfrac{896}{7}$ = 128.

Hence, 128 three digit numbers are divisible by 7.

How many multiples of 4 lie between 10 and 250?

Answer

Multiples of 4 between 10 and 250 are :

12, 16, ........, 248.

The above list is an A.P. with first term (a) = 12 and common difference (d) = 16 - 12 = 4.

Given,

Last term of A.P. = 248.

Let nth term be the last term i.e., 248.

∴ a_n = 248

⇒ a + (n - 1)d = 248

⇒ 12 + 4(n - 1) = 248

⇒ 12 + 4n - 4 = 248

⇒ 4n + 8 = 248

⇒ 4n = 240

⇒ n = $\dfrac{240}{4}$ = 60.

Hence, there are 60 multiples of 4 between 10 and 250.

For what value of n, are the nth terms of two APs: 63, 65, 67, ..... and 3, 10, 17, ........ equal?

Answer

Given,

1st AP : 63, 65, 67, .......

For 1st A.P.,

First term (a) = 63 and Common difference (d) = 65 - 63 = 2.

nth term of 1st A.P. :

= a + (n - 1)d

= 63 + 2(n - 1)

= 63 + 2n - 2

= 2n + 61.

2nd A.P. : 3, 10, 17, .........

First term (a₁) = 3 and Common difference (d₁) = 10 - 3 = 7.

nth term of 2nd A.P. :

= a₁ + (n - 1)d₁

= 3 + 7(n - 1)

= 3 + 7n - 7

= 7n - 4.

Since, nth term of both A.P.s are equal.

∴ 2n + 61 = 7n - 4

⇒ 7n - 2n = 61 + 4

⇒ 5n = 65

⇒ n = $\dfrac{65}{5}$ = 13.

Hence, 13th term of both the A.P.'s are equal.

Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Answer

Let the first term of the A.P. be a and common difference be d.

By formula,

a_n = a + (n - 1)d

Given,

⇒ 3rd term = 16

⇒ a₃ = 16

⇒ a + (3 - 1)d = 16

⇒ a + 2d = 16 ..........(1)

Given,

7th term exceeds the 5th term by 12.

⇒ a₇ - a₅ = 12

⇒ a + (7 - 1)d - [a + (5 - 1)d] = 12

⇒ a + 6d - [a + 4d] = 12

⇒ a - a + 6d - 4d = 12

⇒ 2d = 12

⇒ d = $\dfrac{12}{2}$ = 6.

Substituting value of d in equation (1), we get :

⇒ a + 2 × 6 = 16

⇒ a + 12 = 16

⇒ a = 16 - 12 = 4.

A.P. : a, a + d, a + 2d, a + 3d, ........

⇒ 4, 4 + 6, 4 + 2 × 6, 4 + 3 × 6, .....

⇒ 4, 10, 16, 22, ........

Hence, the required A.P. is 4, 10, 16, 22, ........

Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.

Answer

Given,

A.P. : 3, 8, 13, ......, 253.

Reversing the A.P. : 253, .........., 13, 8, 3.

In above A.P.,

First term (a) = 253 and Common difference (d) = 3 - 8 = -5.

20th term = a₂₀

= a + (20 - 1)d

= 253 + 19 × (-5)

= 253 - 95

= 158.

Hence, 20th term from the last term of the AP : 3, 8, 13, . . ., 253 is 158.

The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Answer

Let the first term of the A.P. be a and common difference be d.

By formula,

a_n = a + (n - 1)d

Given,

Sum of the 4th and 8th terms of an AP is 24.

⇒ a₄ + a₈ = 24

⇒ a + (4 - 1)d + a + (8 - 1)d = 24

⇒ a + 3d + a + 7d = 24

⇒ 2a + 10d = 24

⇒ 2(a + 5d) = 24

⇒ a + 5d = $\dfrac{24}{2}$

⇒ a + 5d = 12 .........(1)

Given,

Sum of the 6th and 10th terms is 44.

⇒ a₆ + a₁₀ = 44

⇒ a + (6 - 1)d + a + (10 - 1)d = 44

⇒ a + 5d + a + 9d = 44

⇒ 2a + 14d = 44

⇒ 2(a + 7d) = 44

⇒ a + 7d = $\dfrac{44}{2}$

⇒ a + 7d = 22 .........(2)

Subtracting equation (1) from (2), we get :

⇒ a + 7d - (a + 5d) = 22 - 12

⇒ a - a + 7d - 5d = 10

⇒ 2d = 10

⇒ d = $\dfrac{10}{2}$

⇒ d = 5.

Substituting value of d in equation (1), we get :

⇒ a + 5 × 5 = 12

⇒ a + 25 = 12

⇒ a = 12 - 25 = -13.

A.P. : a, a + d, a + 2d

⇒ -13, -13 + 5, -13 + 2 × 5

⇒ -13, -8, -3

Hence, first three terms of A.P. are -13, -8, -3.

Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000?

Answer

Starting salary = ₹5000

Increment each year = ₹200

₹5000, ₹5200, ₹5400, ............, ₹7000.

The above list of numbers is an A.P. with first term (a) = ₹5000 and common difference (d) = ₹200.

Let it take n years to reach ₹7000.

By formula,

⇒ a_n = a + (n - 1)d

⇒ 7000 = 5000 + 200(n - 1)

⇒ 7000 - 5000 = 200(n - 1)

⇒ 2000 = 200(n - 1)

⇒ n - 1 = $\dfrac{2000}{200}$

⇒ n - 1 = 10

⇒ n = 10 + 1 = 11.

Hence, in 11th year Subba Rao's income will reach ₹ 7000.

Ramkali saved ₹5 in the first week of a year and then increased her weekly savings by ₹1.75. If in the nth week, her weekly savings become ₹20.75, find n.

Answer

Weekly savings :

5, 6.75, 8.5, ..........., 20.75

The above list is an A.P. with first term (a) = 5 and common difference (d) = 6.75 - 5 = 1.75

Given,

In nth week, weekly savings become ₹20.75

∴ a_n = 20.75

⇒ a + (n - 1)d = 20.75

⇒ 5 + (n - 1) × 1.75 = 20.75

⇒ 5 + 1.75n - 1.75 = 20.75

⇒ 1.75n + 3.25 = 20.75

⇒ 1.75n = 20.75 - 3.25

⇒ 1.75n = 17.5

⇒ n = $\dfrac{17.5}{1.75} = 10.$

Hence, n = 10.

Find the sum of the following APs :

(i) 2, 7, 12,........ to 10 terms.

(ii) -37, -33, -29, ....... to 12 terms.

(iii) 0.6, 1.7, 2.8, ........, to 100 terms.

(iv) $\dfrac{1}{15}, \dfrac{1}{12}, \dfrac{1}{10},$ ......, to 11 terms.

Answer

(i) Given,

A.P. : 2, 7, 12,........ to 10 terms.

In the above A.P.,

First term (a) = 2 and Common difference (d) = 7 - 2 = 5.

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow S_{10} = \dfrac{10}{2}[2 \times 2 + (10 - 1) \times 5] \\[1em] = 5 \times [4 + 9 \times 5] \\[1em] = 5 \times [4 + 45] \\[1em] = 5 \times 49 \\[1em] = 245.$

Hence, sum of the above A.P. is 245.

(ii) Given,

A.P. : -37, -33, -29, ....... to 12 terms.

In the above A.P.,

First term (a) = -37 and Common difference (d) = -33 - (-37) = -33 + 37 = 4.

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow S_{12} = \dfrac{12}{2}[2 \times -37 + (12 - 1) \times 4] \\[1em] = 6 \times [-74 + 11 \times 4] \\[1em] = 6 \times [-74 + 44] \\[1em] = 6 \times -30 \\[1em] = -180.$

Hence, sum of the above A.P. is -180.

(iii) Given,

A.P. : 0.6, 1.7, 2.8,........ to 100 terms.

In the above A.P.,

First term (a) = 0.6 and Common difference (d) = 1.7 - 0.6 = 1.1

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow S_{100} = \dfrac{100}{2}[2 \times 0.6 + (100 - 1) \times 1.1] \\[1em] = 50 \times [1.2 + 99 \times 1.1] \\[1em] = 50 \times [1.2 + 108.9] \\[1em] = 50 \times 110.1 \\[1em] = 5505.$

Hence, sum of the above A.P. is 5505.

(iv) Given,

A.P. : $\dfrac{1}{15}, \dfrac{1}{12}, \dfrac{1}{10},$ ......, to 11 terms.

In the above A.P.,

First term (a) = $\dfrac{1}{15}$ and Common difference (d) = $\dfrac{1}{12} - \dfrac{1}{15} = \dfrac{5 - 4}{60} = \dfrac{1}{60}$.

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow S_{11} = \dfrac{11}{2}\Big[2 \times \dfrac{1}{15} + (11 - 1) \times \dfrac{1}{60}\Big] \\[1em] = \dfrac{11}{2} \times \Big[\dfrac{2}{15} + 10 \times \dfrac{1}{60}\Big] \\[1em] = \dfrac{11}{2} \times \Big[\dfrac{2}{15} + \dfrac{1}{6}\Big] \\[1em] = \dfrac{11}{2} \times \Big[\dfrac{4 + 5}{30}\Big] \\[1em] = \dfrac{11}{2} \times \dfrac{9}{30} \\[1em] = \dfrac{99}{60} = \dfrac{33}{20}.$

Hence, sum of the above A.P. is $\dfrac{33}{20}$.

Find the sums given below :

(i) $7 + 10\dfrac{1}{2} + 14 + ....... + 84$

(ii) 34 + 32 + 30 + ....... + 10

(iii) -5 + (-8) + (-11) + ...... + (-230)

Answer

(i) Given,

⇒ $7 + 10\dfrac{1}{2} + 14 + ....... + 84$

⇒ 7 + 10.5 + 14 + ......... + 84

The above sequence is an A.P. with first term (a) = 7 and common difference (d) = 10.5 - 7 = 3.5

Let nth term be 84.

By formula,

a_n = a + (n - 1)d

Substituting values we get :

⇒ 84 = 7 + (n - 1)(3.5)

⇒ 84 = 7 + 3.5n - 3.5

⇒ 84 = 3.5n + 3.5

⇒ 3.5n = 84 - 3.5

⇒ 3.5n = 80.5

⇒ n = $\dfrac{80.5}{3.5}$ = 23.

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow S_{23} = \dfrac{23}{2}[2 \times 7 + (23 - 1) \times 3.5] \\[1em] = \dfrac{23}{2}[14 + 22 \times 3.5] \\[1em] = \dfrac{23}{2}[14 + 77] \\[1em] = \dfrac{23}{2} \times 91 \\[1em] = \dfrac{2093}{2} \\[1em] = 1046\dfrac{1}{2}.$

Hence, sum = $1046\dfrac{1}{2}$.

(ii) Given,

34 + 32 + 30 + ....... + 10

The above sequence is an A.P. with first term (a) = 34 and common difference (d) = 32 - 34 = -2.

Let nth term be 10.

By formula,

a_n = a + (n - 1)d

Substituting values we get :

⇒ 10 = 34 + (n - 1)(-2)

⇒ 10 = 34 - 2n + 2

⇒ 10 = 36 - 2n

⇒ 2n = 36 - 10

⇒ 2n = 26

⇒ n = $\dfrac{26}{2}$ = 13.

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow S_{13} = \dfrac{13}{2}[2 \times 34 + (13 - 1) \times -2] \\[1em] = \dfrac{13}{2}[68 + 12 \times -2] \\[1em] = \dfrac{13}{2}[68 - 24] \\[1em] = \dfrac{13}{2} \times 44 \\[1em] = 13 \times 22 \\[1em] = 286.$

Hence, sum = 286.

(iii) Given,

-5 + (-8) + (-11) + ....... + (-230)

The above sequence is an A.P. with first term (a) = -5 and common difference (d) = -8 - (-5) = -8 + 5 = -3.

Let nth term be -230.

By formula,

a_n = a + (n - 1)d

Substituting values we get :

⇒ -230 = -5 + (n - 1)(-3)

⇒ -230 = -5 - 3n + 3

⇒ -230 = -3n - 2

⇒ 3n = -2 + 230

⇒ 3n = 228

⇒ n = $\dfrac{228}{3}$ = 76.

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow S_{76} = \dfrac{76}{2}[2 \times -5 + (76 - 1) \times -3] \\[1em] = 38 \times [-10 + 75 \times -3] \\[1em] = 38 \times [-10 - 225] \\[1em] = 38 \times -235 \\[1em] = -8930.$

Hence, sum = -8930.

In an A.P., given a = 5, d = 3, a_n = 50, find n and S_n.

Answer

By formula,

a_n = a + (n - 1)d

⇒ 50 = 5 + 3(n - 1)

⇒ 50 - 5 = 3(n - 1)

⇒ 45 = 3(n - 1)

⇒ n - 1 = 15

⇒ n = 1 + 15 = 16.

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow S_{16} = \dfrac{16}{2}[2 \times 5 + (16 - 1) \times 3] \\[1em] = 8 \times [10 + 15 \times 3] \\[1em] = 8 \times [10 + 45] \\[1em] = 8 \times 55 \\[1em] = 440.$

Hence, n = 16 and S_n = 440.

In an A.P., given a = 7, a₁₃ = 35, find d and S₁₃.

Answer

By formula,

a_n = a + (n - 1)d

Given,

⇒ a₁₃ = 35 and a = 7.

⇒ 7 + (13 - 1)d = 35

⇒ 7 + 12d = 35

⇒ 12d = 28

⇒ d = $\dfrac{28}{12} = \dfrac{7}{3}$

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow S_{13} = \dfrac{13}{2}[2 \times 7 + (13 - 1) \times \dfrac{7}{3}] \\[1em] = \dfrac{13}{2} \times [14 + 12 \times \dfrac{7}{3}] \\[1em] = \dfrac{13}{2} \times [14 + 28] \\[1em] = \dfrac{13}{2} \times 42 \\[1em] = 13 \times 21 \\[1em] = 273.$

Hence, d = $\dfrac{7}{3}$ and S₁₃ = 273.

In an A.P., given a₁₂ = 37, d = 3, find a and S₁₂.

Answer

By formula,

a_n = a + (n - 1)d

Given,

⇒ d = 3

⇒ a₁₂ = 37

⇒ a + 3(12 - 1) = 37

⇒ a + 3(11) = 37

⇒ a + 33 = 37

⇒ a = 4.

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow S_{12} = \dfrac{12}{2}[2 \times 4 + (12 - 1) \times 3] \\[1em] = 6 \times [8 + 11 \times 3] \\[1em] = 6 \times [8 + 33] \\[1em] = 6 \times 41 \\[1em] = 246.$

Hence, a = 4 and S₁₂ = 246.

In an A.P., given a₃ = 15, S₁₀ = 125, find d and a₁₀.

Answer

By formula,

a_n = a + (n - 1)d

Given,

⇒ a₃ = 15

⇒ a + (3 - 1)d = 15

⇒ a + 2d = 15

⇒ a = 15 - 2d .........(1)

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Given,

⇒ S₁₀ = 125

$\Rightarrow \dfrac{10}{2}[2 \times a + (10 - 1)d] = 125 \\[1em] \Rightarrow 5[2a + 9d] = 125 \\[1em] \Rightarrow 2a + 9d = \dfrac{125}{5} \\[1em] \Rightarrow 2a + 9d = 25$

Substituting value of a from equation (1) in above equation :

$\Rightarrow 2(15 - 2d) + 9d = 25 \\[1em] \Rightarrow 30 - 4d + 9d = 25 \\[1em] \Rightarrow 5d = 25 - 30 \\[1em] \Rightarrow 5d = -5 \\[1em] \Rightarrow d = \dfrac{-5}{5} = -1.$

Substituting value of d in equation (1), we get :

⇒ a = 15 - 2d = 15 - 2(-1) = 15 + 2 = 17.

a₁₀ = 17 + (10 - 1)(-1) = 17 - 9 = 8.

Hence, d = -1 and a₁₀ = 8.

In an A.P., given d = 5, S₉ = 75, find a and a₉.

Answer

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Given,

S₉ = 75

$\Rightarrow 75 = \dfrac{9}{2}[2 \times a + (9 - 1) \times 5] \\[1em] \Rightarrow \dfrac{75 \times 2}{9} = 2a + 8 \times 5 \\[1em] \Rightarrow \dfrac{150}{9} = 2a + 40 \\[1em] \Rightarrow 2a + 40 = \dfrac{50}{3} \\[1em] \Rightarrow 2a = \dfrac{50}{3} - 40 \\[1em] \Rightarrow 2a = \dfrac{50- 120}{3} \\[1em] \Rightarrow a = -\dfrac{70}{6} = -\dfrac{35}{3}.$

By formula,

a_n = a + (n - 1)d

Substituting values we get :

$\Rightarrow a_9 = -\dfrac{35}{3} + (9 - 1) \times 5 \\[1em] = -\dfrac{35}{3} + 8 \times 5 \\[1em] = -\dfrac{35}{3} + 40 \\[1em] = \dfrac{-35 + 120}{3} \\[1em] = \dfrac{85}{3}.$

Hence, a = $-\dfrac{35}{3} \text{ and } a_9 = \dfrac{85}{3}$.

In an A.P., given a = 2, d = 8, S_n = 90, find n and a_n.

Answer

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Given,

S_n = 90

$\Rightarrow \dfrac{n}{2}[2 \times 2 + (n - 1) \times 8] = 90 \\[1em] \Rightarrow n[4 + 8n - 8] = 180 \\[1em] \Rightarrow n[8n - 4] = 180 \\[1em] \Rightarrow 8n^2 - 4n = 180 \\[1em] \Rightarrow 4(2n^2 - n) = 180 \\[1em] \Rightarrow 2n^2 - n = \dfrac{180}{4} \\[1em] \Rightarrow 2n^2 - n = 45 \\[1em] \Rightarrow 2n^2 - n - 45 = 0 \\[1em] \Rightarrow 2n^2 - 10n + 9n - 45 = 0 \\[1em] \Rightarrow 2n(n - 5) + 9(n - 5) = 0 \\[1em] \Rightarrow (2n + 9)(n - 5) = 0 \\[1em] \Rightarrow 2n + 9 = 0 \text{ or } n - 5 = 0 \\[1em] \Rightarrow 2n = -9 \text{ or } n = 5 \\[1em] \Rightarrow n = -\dfrac{9}{2} \text{ or } n = 5.$

Since, no. of terms cannot be negative.

∴ n = 5.

By formula,

a_n = a + (n - 1)d

a₅ = 2 + (5 - 1) × 8

= 2 + 4 × 8

= 2 + 32 = 34.

Hence, n = 5 and a_n = 34.

In an A.P., given a = 8, a_n = 62, S_n = 210, find n and d.

Answer

Given,

a_n = 62

By formula,

a_n = a + (n - 1)d

Substituting values we get :

⇒ 62 = 8 + (n - 1)d

⇒ (n - 1)d = 54 ........(1)

Given,

S_n = 210

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow \dfrac{n}{2}[2 \times 8 + (n - 1)d] = 210 \\[1em] \Rightarrow \dfrac{n}{2}[16 + 54] = 210 \\[1em] \Rightarrow \dfrac{n}{2} \times 70 = 210 \\[1em] \Rightarrow 35n = 210 \\[1em] \Rightarrow n = \dfrac{210}{35} = 6.$

Substituting value of n in equation (1), we get :

⇒ (6 - 1)d = 54

⇒ 5d = 54

⇒ d = $\dfrac{54}{5}$.

Hence, n = 6 and d = $\dfrac{54}{5}$.

In an A.P., given a_n = 4, d = 2, S_n = -14, find n and a.

Answer

Given,

a_n = 4

By formula,

a_n = a + (n - 1)d

Substituting values we get :

⇒ 4 = a + 2(n - 1)

⇒ 4 = a + 2n - 2

⇒ a + 2n = 4 + 2

⇒ a + 2n = 6

⇒ a = 6 - 2n ........(1)

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Given,

S_n = -14

Substituting values we get :

$\Rightarrow \dfrac{n}{2}[2 \times (6 - 2n) + 2(n - 1)] = -14 \\[1em] \Rightarrow \dfrac{n}{2}[12 - 4n + 2n - 2] = -14 \\[1em] \Rightarrow \dfrac{n}{2}[10 - 2n] = -14 \\[1em] \Rightarrow 10n - 2n^2 = -28 \\[1em] \Rightarrow 2n^2 - 10n - 28 = 0 \\[1em] \Rightarrow 2(n^2 - 5n - 14) = 0 \\[1em] \Rightarrow n^2 - 5n - 14 = 0 \\[1em] \Rightarrow n^2 - 7n + 2n - 14 = 0 \\[1em] \Rightarrow n(n - 7) + 2(n - 7) = 0 \\[1em] \Rightarrow (n + 2)(n - 7) = 0 \\[1em] \Rightarrow n + 2 = 0 \text{ or } n - 7 = 0 \\[1em] \Rightarrow n = -2 \text{ or } n = 7.$

Since, no. of term cannot be negative.

∴ n = 7.

Substituting value of n in equation (1), we get :

⇒ a = 6 - 2n = 6 - 2 × 7 = 6 - 14 = -8.

Hence, a = -8 and n = 7.

In an A.P., given a = 3, n = 8, S = 192, find d.

Answer

Given,

a = 3, n = 8.

By formula,

Sum of n terms = $\dfrac{n}{2}[2a + (n - 1)d]$

S = 192

Substituting values we get :

$\Rightarrow \dfrac{8}{2}[2 \times 3 + (8 - 1) \times d] = 192 \\[1em] \Rightarrow 4[6 + 7d] = 192 \\[1em] \Rightarrow 6 + 7d = \dfrac{192}{4} \\[1em] \Rightarrow 6 + 7d = 48 \\[1em] \Rightarrow 7d = 42 \\[1em] \Rightarrow d = \dfrac{42}{7} = 6.$

Hence, d = 6.

In an A.P., given l = 28, S = 144, and there are total 9 terms. Find a.

Answer

By formula,

Sum of n terms = $\dfrac{n}{2}$ [First term + Last term]

Given,

n = 9 and l = 28.

S₉ = 144

Substituting values we get :

$\Rightarrow \dfrac{9}{2}[a + 28] = 144 \\[1em] \Rightarrow a + 28 = \dfrac{144 \times 2}{9} \\[1em] \Rightarrow a + 28 = 16 \times 2 \\[1em] \Rightarrow a + 28 = 32 \\[1em] \Rightarrow a = 32 - 28 = 4.$

Hence, a = 4.

How many terms of the AP : 9, 17, 25,....... must be taken to give a sum of 636?

Answer

Given,

A.P. : 9, 17, 25,.......

First term (a) = 9 and common difference (d) = 17 - 9 = 8.

Let sum of n terms of the above A.P. be 636.

∴ S_n = 636

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow 636 = \dfrac{n}{2}[2 \times 9 + (n - 1) \times 8] \\[1em] \Rightarrow 636 = \dfrac{n}{2}[18 + (n - 1) \times 8] \\[1em]\Rightarrow 636 = \dfrac{n}{2}[18 + 8n - 8] \\[1em] \Rightarrow 636 = \dfrac{n}{2}[8n + 10] \\[1em] \Rightarrow 636 = \dfrac{8n^2 + 10n}{2} \\[1em] \Rightarrow 8n^2 + 10n = 1272 \\[1em] \Rightarrow 2(4n^2 + 5n) = 1272 \\[1em] \Rightarrow 4n^2 + 5n = \dfrac{1272}{2} \\[1em] \Rightarrow 4n^2 + 5n = 636 \\[1em] \Rightarrow 4n^2 + 5n - 636 = 0$

Comparing equation,

4n² + 5n - 636 = 0 with ax² + bx + c = 0, we get :

a = 4, b = 5 and c = -636.

By formula,

n = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values we get :

$\Rightarrow n = \dfrac{-5 \pm \sqrt{5^2 - 4 \times 4 \times -636}}{2 \times 4} \\[1em] = \dfrac{-5 \pm \sqrt{25 + 10176}}{8} \\[1em] = \dfrac{-5 \pm \sqrt{10201}}{8} \\[1em] = \dfrac{-5 \pm 101}{8} \\[1em] = \dfrac{-5 + 101}{8}, \dfrac{-5 - 101}{8} \\[1em] = \dfrac{96}{8}, -\dfrac{106}{8} \\[1em] = 12, -\dfrac{53}{4}$

Since, no. of term cannot be negative.

∴ n = 12.

Hence, sum of 12 terms of A.P. = 636.

The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Answer

Given,

First term (a) = 5

Last term (l) = 45

Let number of terms be n and common difference be d.

By formula,

S = $\dfrac{n}{2}[a + l]$

Substituting values, we get :

$\Rightarrow 400 = \dfrac{n}{2}[5 + 45] \\[1em] \Rightarrow 400 = \dfrac{n}{2} \times 50 \\[1em] \Rightarrow 400 = 25n \\[1em] \Rightarrow n = \dfrac{400}{25} \\[1em] \Rightarrow n = 16.$

Since, nth term is the last term.

⇒ a_n = 45

⇒ a + (n - 1)d = 45

⇒ 5 + (16 - 1)d = 45

⇒ 5 + 15d = 45

⇒ 15d = 40

⇒ d = $\dfrac{40}{15} = \dfrac{8}{3}$.

Hence, no. of terms = 16 and common difference = $\dfrac{8}{3}$.

The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Answer

Let there be n terms in the A.P.

Given,

First term (a) = 17

Last term (l) = a_n = 350

Common difference (d) = 9

By formula,

a_n = a + (n - 1)d

Substituting values we get :

⇒ 350 = 17 + 9(n - 1)

⇒ 350 = 17 + 9n - 9

⇒ 350 = 9n + 8

⇒ 9n = 350 - 8

⇒ 9n = 342

⇒ n = $\dfrac{342}{9}$ = 38.

By formula,

Sum of A.P. (S) = $\dfrac{n}{2}(a + l)$

Substituting values we get :

$\text{Sum of first 38 terms} = \dfrac{38}{2}(17 + 350) \\[1em] = 19 \times 367 \\[1em] = 6973.$

Hence, sum of A.P. = 6973 and number of terms = 38.

Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Answer

Let first term of A.P. be a.

Given,

Common difference (d) = 7

By formula,

a_n = a + (n - 1)d

Given,

⇒ a₂₂ = 149

⇒ a + 7(22 - 1) = 149

⇒ a + 7 × 21 = 149

⇒ a + 147 = 149

⇒ a = 149 - 147 = 2.

Last term (l) = 22nd term = 149

By formula,

Sum of A.P. (S) = $\dfrac{n}{2}(a + l)$

Substituting values we get :

$\text{Sum of first 22 terms} = \dfrac{22}{2}(2 + 149) \\[1em] = 11 \times 151 \\[1em] = 1661.$

Hence, sum of first 22 terms = 1661.

Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Answer

Let first term of A.P. be a and common difference be d.

By formula,

a_n = a + (n - 1)d

Given,

2nd term = 14

⇒ a₂ = 14

⇒ a + (2 - 1)d = 14

⇒ a + d = 14 ............(1)

3rd term = 18

⇒ a₃ = 18

⇒ a + (3 - 1)d = 18

⇒ a + 2d = 18 ............(2)

Subtracting equation (1) from (2), we get :

⇒ a + 2d - (a + d) = 18 - 14

⇒ a - a + 2d - d = 4

⇒ d = 4.

Substituting value of d in equation (1), we get :

⇒ a + 4 = 14

⇒ a = 10.

Last term (l) = 51st term = a₅₁

= a + (51 - 1)d

= 10 + 50 × 4

= 10 + 200

= 210.

By formula,

Sum (S) = $\dfrac{n}{2}[a + l]$

Substituting values we get :

$\text{Sum of first 51 terms} = \dfrac{51}{2}[10 + 210] \\[1em] = \dfrac{51}{2} \times 220 \\[1em] = 51 \times 110 \\[1em] = 5610.$

Hence, sum of first 51 terms = 5610.

If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Answer

Let first term be a and common difference be d.

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Given,

Sum of first 7 terms = 49

$\therefore S_7 = \dfrac{7}{2}[2 \times a + (7 - 1)d] \\[1em] \Rightarrow 49 = \dfrac{7}{2}[2a + 6d] \\[1em] \Rightarrow 49 = \dfrac{7}{2} \times 2[a + 3d] \\[1em] \Rightarrow \dfrac{49}{7} = a + 3d \\[1em] \Rightarrow a + 3d = 7 \text{ .......(1)}.$

Given,

Sum of first 17 terms = 289

$\therefore S_{17} = \dfrac{17}{2}[2 \times a + (17 - 1)d] \\[1em] \Rightarrow 289 = \dfrac{17}{2}[2a + 16d] \\[1em] \Rightarrow 289 = \dfrac{17}{2} \times 2[a + 8d] \\[1em] \Rightarrow \dfrac{289}{17} = a + 8d \\[1em] \Rightarrow a + 8d = 17 \text{ .......(2)}.$

Subtracting equation (1) from (2), we get :

⇒ a + 8d - (a + 3d) = 17 - 7

⇒ a - a + 8d - 3d = 10

⇒ 5d = 10

⇒ d = $\dfrac{10}{5}$ = 2.

Substituting value of d in equation (1), we get :

⇒ a + 3 × 2 = 7

⇒ a + 6 = 7

⇒ a = 1.

$S_n = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] = \dfrac{n}{2}[2 \times 1 + 2(n - 1)] \\[1em] = \dfrac{n}{2}[2 + 2n - 2] \\[1em] = \dfrac{n}{2} \times 2n \\[1em] = n^2.$

Hence, sum of first n terms = n².

Show that a₁, a₂,......., a_n,........ form an AP where a_n is defined as below :

(i) a_n = 3 + 4n

(ii) a_n = 9 - 5n

Also find the sum of the first 15 terms in each case.

Answer

(i) Given,

⇒ a_n = 3 + 4n

⇒ a₁ = 3 + 4 × 1 = 3 + 4 = 7

⇒ a₂ = 3 + 4 × 2 = 3 + 8 = 11

⇒ a₃ = 3 + 4 × 3 = 3 + 12 = 15

a₂ - a₁ = 11 - 7 = 4,

a₃ - a₂ = 15 - 11 = 4.

Since, a₃ - a₂ = a₂ - a₁.

Hence, it is an A.P. with common difference = 4 and first term (a₁) = 7.

a₁₅ = 3 + 4 × 15 [∵ a_n = 3 + 4n]

= 3 + 60 = 63.

By formula,

Sum of A.P. = $\dfrac{n}{2}$ [First term + Last term]

Substituting values we get :

$S_{15} = \dfrac{15}{2}[7 + 63] \\[1em] = \dfrac{15}{2} \times 70 \\[1em] = 15 \times 35 \\[1em] = 525.$

Hence, sum of first 15 terms = 525.

(ii) Given,

⇒ a_n = 9 - 5n

⇒ a₁ = 9 - 5 × 1 = 9 - 5 = 4

⇒ a₂ = 9 - 5 × 2 = 9 - 10 = -1

⇒ a₃ = 9 - 5 × 3 = 9 - 15 = -6

a₂ - a₁ = -1 - 4 = -5,

a₃ - a₂ = -6 - (-1) = -5.

Since, a₃ - a₂ = a₂ - a₁.

Hence, it is an A.P. with common difference = -5 and first term (a₁) = 4.

a₁₅ = 9 - (5 x 15) [∵ a_n = 9 - 5n]

= 9 - 75

= -66.

By formula,

Sum of A.P. = $\dfrac{n}{2}$ [First term + Last term]

Substituting values we get :

$S_{15} = \dfrac{15}{2}[4 + (-66)] \\[1em] = \dfrac{15}{2} \times -62 \\[1em] = 15 \times -31 \\[1em] = -465.$

Hence, sum of first 15 terms = -465.

If the sum of the first n terms of an AP is 4n – n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

Answer

Given,

S_n = 4n - n²

First term = S₁

= 4(1) - 1²

= 4 - 1 = 3.

S₂ = 4(2) - 2²

= 8 - 4

= 4.

Second term = S₂ - S₁

= 4 - 3 = 1.

Third term = S₃ - S₂

= [4(3) - 3²] - 4

= [12 - 9] - 4

= -1.

Tenth term = S₁₀ - S₉

= [4(10) - 10²] - [4(9) - 9²]

= [40 - 100] - [36 - 81]

= [-60] - [-45]

= -60 + 45

= -15.

nth term = S_n - S_{n - 1}

= [4n - n²] - [4(n - 1) - (n - 1)²]

= [4n - n²] - [4n - 4 - (n² + 1 - 2n)]

= [4n - n²] - [4n - 4 - n² - 1 + 2n]

= 4n - 4n - n² + n² + 4 + 1 - 2n

= 5 - 2n.

Hence, first term = 3, sum upto 2 terms = 4, second term = 1, third term = -1, tenth term = -15 and nth term = 5 - 2n.

Find the sum of the first 40 positive integers divisible by 6.

Answer

List of positive integers divisible by 6 are :

6, 12, 18, ........

The above list is an A.P. with first term (a) = 6 and common difference (d) = 12 - 6 = 6.

By formula,

Sum of first n terms = S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$S_{40} = \dfrac{40}{2}[2 \times 6 + (40 - 1) \times 6] \\[1em] = 20 \times [12 + 39 \times 6] \\[1em] = 20 \times [12 + 234] \\[1em] = 20 \times 246 \\[1em] = 4920.$

Hence, sum of first 40 positive integers divisible by 6 = 4920.

Find the sum of the first 15 multiples of 8.

Answer

List of multiples of 8 are :

8, 16, 24, 32, ........

The above list is an A.P. with first term (a) = 8 and common difference (d) = 16 - 8 = 8.

By formula,

Sum of first n terms = S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$S_{15} = \dfrac{15}{2}[2 \times 8 + (15 - 1) \times 8] \\[1em] = \dfrac{15}{2}[16 + 14 \times 8] \\[1em] = \dfrac{15}{2} \times [16 + 112] \\[1em] = \dfrac{15}{2} \times 128 \\[1em] = 15 \times 64 \\[1em] = 960.$

Hence, sum of first 15 multiples of 8 = 960.

Find the sum of the odd numbers between 0 and 50.

Answer

Sum of odd numbers between 0 and 50 are :

1, 3, 5, ........, 49.

The above list is an A.P. with first term (a) = 1 and common difference (d) = 3 - 1 = 2.

Let 49 be the nth term.

⇒ a_n = 49

⇒ a + (n - 1)d = 49

⇒ 1 + 2(n - 1) = 49

⇒ 2(n - 1) = 48

⇒ n - 1 = $\dfrac{48}{2}$

⇒ n - 1 = 24

⇒ n = 25.

By formula,

Sum of an A.P. (S) = $\dfrac{n}{2}$ [First term + Last term]

Substituting values we get :

$S = \dfrac{25}{2}[1 + 49] \\[1em] = \dfrac{25}{2} \times 50 \\[1em] = 25 \times 25 \\[1em] = 625.$

Hence, sum of the odd numbers between 0 and 50 = 625.

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, etc., the penalty for each succeeding day being ₹50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?

Answer

Since, work has been delayed by 30 days.

Total Penalty (in ₹) : 200 + 250 + 300 + ......... + upto 30 terms

The above sequence is an A.P. with first term (a) = ₹ 200 and common difference = ₹ 50.

By formula,

Sum upto n terms = S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$S_{30} = \dfrac{30}{2}[2 \times 200 + (30 - 1) \times 50] \\[1em] = 15 \times [400 + 29 \times 50] \\[1em] = 15 \times [400 + 1450] \\[1em] = 15 \times 1850 \\[1em] = \text{₹ 27750}.$

Hence, total penalty = ₹ 27750.

A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.

Answer

Let value of highest price be ₹ x. Given, each prize is ₹ 20 less than preceding prize.

Prizes : x, x - 20, x - 40, ........ upto 7 terms.

The above list is an A.P. with first term (a) = x and common difference = -20.

Total prize = ₹ 700.

By formula,

Sum upto n terms = S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow S_7 = \dfrac{7}{2}[2 \times x + (7 - 1) \times -20] \\[1em] \Rightarrow 700 = \dfrac{7}{2}\times [2x - 120] \\[1em] \Rightarrow 2x - 120 = \dfrac{700 \times 2}{7} \\[1em] \Rightarrow 2x - 120 = 200 \\[1em] \Rightarrow 2(x - 60) = 200 \\[1em] \Rightarrow x - 60 = 100 \\[1em] \Rightarrow x = \text{₹ 160}.$

Next terms are :

x - ₹ 20 = ₹ 160 - ₹ 20 = ₹140,

₹ 140 - ₹ 20 = ₹120,

₹ 120 - ₹ 20 = ₹100,

₹ 100 - ₹ 20 = ₹80,

₹ 80 - ₹ 20 = ₹60,

₹ 60 - ₹ 20 = ₹40.

Hence, prizes are : ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40.

In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

Answer

For Class 1 each section will plant 1 tree and there are 3 sections so total 3 plants will be planted by class 1st.

For Class 2nd each section will plant 2 trees and there are 3 sections so total 6 plants will be planted by class 2nd.

Similarly for class 3rd total trees will be 9.

So,

Total trees = 3 + 6 + 9 + ...... + upto 12 terms.

The above list is an A.P. with first term (a) = 3 and common difference = 6 - 3 = 3.

By formula,

Sum upto n terms = S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$S_{12} = \dfrac{12}{2}[2 \times 3 + (12 - 1) \times 3] \\[1em] = 6[6 + 11 \times 3] \\[1em] = 6 \times [6 + 33] \\[1em] = 6 \times 39 \\[1em] = 234.$

Hence, total trees planted = 234.

A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ....... as shown in the fig. below. What is the total length of such a spiral made up of thirteen consecutive semicircles?

$(\text{Take } \pi = \dfrac{22}{7})$

Answer

Circumference of semicircle = πr

Circumference of first circle = 0.5 π

Circumference of second circle = 1π = π

Circumference of third circle = 1.5π

Total length = 0.5π + π + 1.5π ......... upto 13 terms

The above list is an A.P. with first term (a) = 0.5π and common difference (d) = π - 0.5π = 0.5π

By formula,

Sum upto n terms = S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow S_{13} = \dfrac{13}{2}[2 \times 0.5π + (13 - 1) \times 0.5π] \\[1em] = \dfrac{13}{2}[π + 12 \times 0.5π] \\[1em] = \dfrac{13}{2}[π + 6π] \\[1em] = \dfrac{13}{2} \times 7π \\[1em] = 13 \times 3.5π \\[1em] = 13 \times 3.5 \times \dfrac{22}{7} \\[1em] = 13 \times 0.5 \times 22 \\[1em] = 13 \times 11 \\[1em] = 143 \text{ cm}.$

Hence, total length of spiral = 143 cm.

200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

Answer

Given,

20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on.

Let n rows be required to stack 200 logs.

Total logs = 20 + 19 + 18 + ......... upto n terms.

In above sequence, first term (a) = 20 and common difference (d) = 19 - 20 = -1.

By formula,

Sum upto n terms = S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow 200 = \dfrac{n}{2}[2 \times 20 + (n - 1) \times (-1)] \\[1em] \Rightarrow 200 = \dfrac{n}{2}[40 - n + 1] \\[1em] \Rightarrow 400 = n[41 - n] \\[1em] \Rightarrow 41n - n^2 = 400 \\[1em] \Rightarrow n^2 - 41n + 400 = 0 \\[1em] \Rightarrow n^2 - 16n - 25n + 400 = 0 \\[1em] \Rightarrow n(n - 16) - 25(n - 16) = 0 \\[1em] \Rightarrow (n - 25)(n - 16) = 0 \\[1em] \Rightarrow n - 25 = 0 \text{ or } n - 16 = 0 \\[1em] \Rightarrow n = 25 \text{ or } n = 16.$