Question

how many terms of the seris 64+32+16+....must be taken so that sum may be 127.5?​

Answer 1

Answer:

hope my answer helps you.....

Answer 2

The given series are in geometric progression,

$64+32+16+.....$

Common ratio$r=\frac{ar}{a}$

$=\frac{32}{64}\\=\frac{1}{2}$

The value of r is less than 1.

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

$S_n=127.5, a=64, r=\frac{1}{2}, n=?$

$127.5=\frac{64(1-{\frac{1}{2}}^n)}{1-\frac{1}{2}}\\\frac{255}{2}=\frac{64(1-{\frac{1}{2}}^n)}{\frac{1}{2}}\\\frac{255}{2}=64\times 2(1-{\frac{1}{2}}^n)}\\\frac{255}{2\times 128}=(1-{\frac{1}{2}}^n)}\\\frac{255}{256}=(1-{\frac{1}{2}}^n)}\\{\frac{1}{2}}^n=1-\frac{255}{256}\\{\frac{1}{2}}^n=\frac{1}{256}\\{\frac{1}{2}}^n=\frac{1}{2^8}\\{\frac{1}{2}}^n={\frac{1}{2}}^8$

Hence, $n=8$

8 terms are required of the given series whose sum is 127.5.