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  • Let  be the nth term of a G.P. of positive terms. If <img alt="\sum_{n=1}^{100}a_{2n+1}=200\: \: and\: \: \sum_{n=1}^{100}a_{2n}=100,\: \:

Let a_{n} be the nth term of a G.P. of positive terms. If \sum_{n=1}^{100}a_{2n+1}=200\: \: and\: \: \sum_{n=1}^{100}a_{2n}=100,\: \: then\: \sum_{n=1}^{200}a_n is equal to :   
Option: 1 300
Option: 2 175
Option: 3 225
Option: 4 150
 

Answers (5)

best_answer

Sum of n-term of a GP

Let Sn be the sum of n terms of the G.P. with the first term ‘a’ and common ratio ‘r’. Then 

S_n=a(\frac{1-r^n}{1-r})

The above formula does not hold for r = 1, For r = 1 the sum of n terms of the G.P. is S_n = na.

 

Now,

\\\sum_{n=1}^{100} a_{2 n+1}=a_{3}+a_{5}+\ldots \ldots a_{201}=200 \Rightarrow \frac{\operatorname{ar}^{2}\left(r^{200}-1\right)}{r^{2}-1}=200\\\sum_{n=1}^{100} a_{2 n}=a_{2}+a_{4}+\ldots \ldots . a_{200}=100=\frac{\operatorname{ar}\left(r^{200}-1\right)}{r^{2}-1}=100

On dividing above equations, r = 2

Adding both equations,

\\\Rightarrow \mathrm{a}_{2}+\mathrm{a}_{3}+\ldots \ldots \ldots \ldots \mathrm{a}_{200}+\mathrm{a}_{201}=300\\\Rightarrow r\left(a_{1}+\ldots \ldots \ldots \ldots a_{200}\right)=300\\\sum_{n=1}^{200} a_{n}=\frac{300}{r}=150

Correct Option 4

Posted by

avinash.dongre

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225

Posted by

Anakha

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150

Posted by

L.venkateswarlu Laddagiri

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225

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Posted by

Rohit Mangal

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150

Posted by

Nalla mahalakshmi

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