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.

(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 $\dfrac{33}{20}$.