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7.Find sum up to infinite terms of the series 1+ (3)/(5)+ (6)/(5^(2))+ (10)/(5^(3))+ (15)/(5^(4))+ (21)/(5^(5))+...... (a) (5)/(4) (b) (25)/(16) (c) (125)/(64) (d) (100)/(3)


Prasanth 14th Oct, 2020
Answer (1)
Shakti Swarupa Bhanja 6th Nov, 2020

Hello!

I tried solving the question and the solution comes out to be option (c) 125/64 .

I am explaining the solution below:-

S = 1+ (3/5)+ (6/5^2)+ (10/5^3)+ (15/5^4)+......

=>S/5 = (1/5)+ (3/5^2)+ (6/5^3)+ (10/5^4)+.....

=>4S/5= (1/5)+ (2/5)+ (3/5^2)+ (4/5^3)+ (5/5^3)+....

=>4S/5^2= (1/5)+ (2/5^2)+ (3/5^3)+ (4/5^3)+....

We can take,

(4S/5) - (4S/25)= 1+ (1/5)+ (1/5^2)+ (1/5^3)+.....

=>(16S/25)= 1+ (1/5)+ (1/5^2)+......[Infinite G.P]

We have, a=1 and r=(1/5).

So, S=a/(1-r) =1/(1-1/5) = 1/(4/5)

=>16S/25= 5/4 => S= 125/64.

Hope this cleared your confusion!

Happy learning:)

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