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Mathematics

Find the sum of :

(i) 20 terms of the series 2 + 6 + 18 + …

(ii) 10 terms of the series 1 + 3\sqrt{3} + 3 + ….

(iii) 6 terms of the G.P. 1, −23,49,-\dfrac{2}{3}, \dfrac{4}{9}, ….

(iv) 5 terms and n terms of the series 1 + 23+49+....\dfrac{2}{3} + \dfrac{4}{9} + ….

AP GP

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Answer

(i) Sum = 2 + 6 + 18 + ….. + (20th term)

The above list of numbers is a G.P. with first term a = 2 and common ratio = r = 62=3.\dfrac{6}{2} = 3.

By formula,

Sn=a(rn−1)r−1∴S20=2((3)20−1)3−1=2(320−1)2=320−1.Sn = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \therefore S{20} = \dfrac{2((3)^{20} - 1)}{3 - 1} \\[1em] = \dfrac{2(3^{20} - 1)}{2} \\[1em] = 3^{20} - 1.

Hence, the sum of the series is 320 - 1.

(ii) Sum = 1 + 3\sqrt{3} + 3 + …. + (10th term)

The above list of numbers is a G.P. with first term a = 1 and common ratio = r = 3.\sqrt{3}.

By formula,

Sn=a(rn−1)r−1∴S10=1((3)10−1)3−1=((3)10/2−1)3−1=35−13−1Sn = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \therefore S{10} = \dfrac{1((\sqrt{3})^{10} - 1)}{\sqrt{3} - 1} \\[1em] = \dfrac{((3)^{10/2} - 1)}{\sqrt{3} - 1} \\[1em] = \dfrac{3^5 - 1}{\sqrt{3} - 1}

Multiplying numerator and denominator by 3\sqrt{3} + 1,

=35−13−1×3+13+1=242(3+1)(3−1)(3+1)=242(3+1)3+3−3−1=242(3+1)2=121(3+1).= \dfrac{3^5 - 1}{\sqrt{3} - 1} \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1} \\[1em] = \dfrac{242(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \\[1em] = \dfrac{242(\sqrt{3} + 1)}{3 + \sqrt{3} - \sqrt{3} - 1} \\[1em] = \dfrac{242(\sqrt{3} + 1)}{2} \\[1em] = 121(\sqrt{3} + 1).

Hence, the sum of the series is 121(3+1)121(\sqrt{3} + 1).

(iii) Sum = 1, −23,49,-\dfrac{2}{3}, \dfrac{4}{9}, ….(6th term)

The above list of numbers is a G.P. with first term a = 1 and common ratio = r = −23.-\dfrac{2}{3}.

By formula,

Sn=a(rn−1)r−1∴S6=1[(−23)6−1]−23−1=2636−1−2−33=2636−1−53=3(26−36)36×−5=64−72935×−5=−665−1215Sn = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \therefore S{6} = \dfrac{1\Big[\Big(-\dfrac{2}{3}\Big)^6 - 1\Big]}{-\dfrac{2}{3} - 1} \\[1em] = \dfrac{\dfrac{2^6}{3^6} - 1}{\dfrac{-2 -3}{3}} \\[1em] = \dfrac{\dfrac{2^6}{3^6} - 1}{-\dfrac{5}{3}} \\[1em] = \dfrac{3(2^6 - 3^6)}{3^6 \times -5} \\[1em] = \dfrac{64 - 729}{3^5 \times -5} \\[1em] = \dfrac{-665}{-1215}

On dividing numerator and denominator by -5 we get,

=133243.= \dfrac{133}{243}.

Hence, the sum of the series is 133243\dfrac{133}{243}.

(iv) Sum upto 5 terms = 1 + 23+49+....\dfrac{2}{3} + \dfrac{4}{9} + …. + 5th term

The above list of numbers is a G.P. with first term a = 1 and common ratio = r = 23.\dfrac{2}{3}.

By formula,

Sn=a(rn−1)r−1∴S5=1[(23)5−1]23−1=2535−12−33=2535−1−13=3(25−35)35×−1=32−24334×−1=−211−81Sn = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \therefore S5 = \dfrac{1\Big[\Big(\dfrac{2}{3}\Big)^5 - 1\Big]}{\dfrac{2}{3} - 1} \\[1em] = \dfrac{\dfrac{2^5}{3^5} - 1}{\dfrac{2 -3}{3}} \\[1em] = \dfrac{\dfrac{2^5}{3^5} - 1}{-\dfrac{1}{3}} \\[1em] = \dfrac{3(2^5 - 3^5)}{3^5 \times -1} \\[1em] = \dfrac{32 - 243}{3^4 \times -1} \\[1em] = \dfrac{-211}{-81}

On dividing numerator and denominator by -1 we get,

=21181= \dfrac{211}{81}

Sum of n terms = Sn=1[(23)n−1]23−1S_n = \dfrac{1\Big[\Big(\dfrac{2}{3}\Big)^n - 1\Big]}{\dfrac{2}{3} - 1}

=(23)n−12−33=3[(23)n−1]−1=−3[(23)n−1]=3[1−(23)n].= \dfrac{\Big(\dfrac{2}{3}\Big)^n - 1}{\dfrac{2 - 3}{3}} \\[1em] = \dfrac{3\Big[\Big(\dfrac{2}{3}\Big)^n - 1\Big]}{-1} \\[1em] = -3\Big[\Big(\dfrac{2}{3}\Big)^n - 1\Big] \\[1em] = 3\Big[1 - \Big(\dfrac{2}{3}\Big)^n\Big].

Hence, the sum of the series upto 5 terms is 21181\dfrac{211}{81} and upto n terms is 3[1−(23)n]3\Big[1 - \Big(\dfrac{2}{3}\Big)^n\Big].

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