Question

if 6 term of gp is -1/32 and 9 term is 1/256 then 11 term =

Answer 1

Hey Friends,
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Let the first term be a

$t_{6} = {ar}^{6 - 1}$
$t_{6} = {ar}^{5}$
$\frac{ - 1}{32} = {ar}^{5}$ --------(i)
$t_{9} = {ar}^{8}$
$\frac{1}{256} = {ar}^{8}$ --------(ii)
By Dividing equ-(ii) by equ-(i)
$\frac{ -32}{256} = \frac{ {ar}^{8} }{ {ar}^{5} }$
$\frac{ - 1}{8} = {r}^{3}$
$r = \frac{ - 1}{2}$
Put the value of r in equ-(i)

$\frac{ - 1}{32} = a( \frac{ - 1}{2}) ^{5}$
$a = 1$
$t_{11} = 1 \times ( \frac{ - 1}{2} ) ^{10}$

$t_{11} = \frac{1}{1024}$

Hope it's help you!!!!
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