Question

Find the G.P. if $S_{10}:S_5=33:32$ and $T_5=64$

Answer 1

We know that the formula for sum of $n$ terms of G.P is $S_n=\frac{a(r^n-1)}{r-1}$, where $a$ is the first term, $r$ is the common ratio.

It is given thatthe ratio of the sum $S_{10}$ and $S_5$ is $33:32$, therefore,

$\frac{S_{10}}{S_5}=\frac{\frac{a(r^{10}-1)}{r-1}}{\frac{a(r^5-1)}{r-1}}\Rightarrow \frac{33}{32}=\frac{r^{10}-1}{r^5-1}\Rightarrow \frac{33}{32}=\frac{(r^5{)}^2-1}{r^5-1}\Rightarrow \frac{33}{32}=\frac{(r^5-1)(r^5+1)}{r^5-1}\Rightarrow \frac{33}{32}=r^5+1\Rightarrow 33=32r^5+32\Rightarrow 32r^5=33-32=1\Rightarrow r^5=\frac{1}{32}\Rightarrow r^5=\frac{1}{2^5}\Rightarrow r^5={\left(\frac{1}{2}\right)}^5\Rightarrow r=\frac{1}{2}$

Now, the formula for the $n$th term of an G.P is $T_n=a{r}^{n-1}$, where $a$ is the first term, $r$ is the common ratio.

We are given that $T_5=64$ where $n=5$ and $r=\frac{1}{2}$, thus,

$T_n=a{r}^{n-1}\Rightarrow T_5=a\times {\left(\frac{1}{2}\right)}^{5-1}\Rightarrow 64=a\times \frac{1}{2^4}\Rightarrow 64=\frac{a}{16}\Rightarrow a=64\times 16=1024$

Now, with $a=1024$ and $r=\frac{1}{2}$, the required terms of G.P are:

$a_1=1024a_2=a_1\times r=1024\times \frac{1}{2}=512a_3=a_2\times r=512\times \frac{1}{2}=256a_4=a_3\times r=256\times \frac{1}{2}=128$

Hence the G.P is $1024,512,256,128,.....$.