Question

Find the value of n for which

$704+\frac{1}{2}\left(704\right)+\frac{1}{4}\left(704\right)+...$ upto n terms = $1984-\frac{1}{2}\left(1984\right)+\frac{1}{4}\left(1984\right)-...$ upto n terms.

Answer 1

For LHS $a=704,r=\frac{1}{2}$

For RHS $a=1984,r=-\frac{1}{2}$

sum of n terms of GP

$S_n=\frac{a(1-rn)}{(1-r)}$

$704\frac{[1–\frac{1}{2}n]}{(1–\frac{1}{2})}=1984\frac{[1–(-\frac{1}{2}\left)n\right]}{(1–(-\frac{1}{2}\left)\right)}$

$704\times 2(1–\frac{1}{2n})=1984\times \frac{2}{3}\times [1–(-1\left)\frac{n}{2n}\right]$

$704\times 6(1–\frac{1}{2n})=1984\times 2[1–(-1\left)\frac{n}{2n}\right]$

$\frac{4224–4224}{2n}=\frac{3968-3968(-1)n}{2n}$

$4224-3968=\frac{4224}{2n}-\frac{3968(-1)n}{2n}$

$256=\frac{4224}{2n}-\frac{3968(-1)n}{2n}$

$128=\frac{2112}{2n}-\frac{1984(-1)n}{2n}$

$128=\frac{(2112+1984)}{2n}$

$2n=\frac{4096}{128}=\frac{1024}{32}=\frac{128}{4}=32$

$2n=32$

$n=5$