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The 2nd and 5th terms of a GP are

Question:

The $2^{\text {nd }}$ and $5^{\text {th }}$ terms of a GP are $\frac{-1}{2}$ and $\frac{1}{16}$ respectively. Find the sum of $n$ terms GP up to 8 terms.

 

Solution:

$2^{\text {nd }}$ term $=a r^{2-1}=a r^{1}$

$5^{\text {th }}$ term $=a r^{5-1}=a r^{4}$

Dividing the $5^{\text {th }}$ term using the $3^{\text {rd }}$ term

$\frac{a r^{4}}{a r}=\frac{\frac{1}{16}}{\frac{-1}{2}}$

$r^{3}=-\frac{1}{8}$

$\therefore r=-\frac{-1}{2}$

∴ a = 1

Sum of a G.P. series is represented by the formula $S_{n}=a \frac{1-r^{n}}{1-r}$ when |r|<1. ‘Sn’ represents the sum of the G.P. series upto nth terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.

n = 8 terms

$\mathrm{S}_{\mathrm{n}}=1 \times \frac{1-\frac{-1^{8}}{2}}{1-\frac{-1}{2}}$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{1-\frac{1}{256}}{\frac{3}{2}}$

$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\frac{255}{256}}{\frac{3}{2}}$

$\therefore \mathrm{S}_{\mathrm{n}}=\frac{170}{256}$

 

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