Question

Find the G.P. if $S_6:S_3=126:1$

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 $S_6$ and $S_3$ is $126:1$, therefore,

$\frac{S_6}{S_3}=\frac{\frac{a(r^6-1)}{r-1}}{\frac{a(r^3-1)}{r-1}}\Rightarrow \frac{126}{1}=\frac{r^6-1}{r^3-1}\Rightarrow 126=\frac{(r^3{)}^2-1}{r^3-1}\Rightarrow 126=\frac{(r^3-1)(r^3+1)}{r^3-1}\Rightarrow 126=r^3+1\Rightarrow r^3=126-1=125\Rightarrow r^3=5^3\Rightarrow r=5$

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 $T_4=125$ where $n=4$ and $r=5$, thus,

$T_n=a{r}^{n-1}\Rightarrow T_4=a\times {\left(5\right)}^{4-1}\Rightarrow 125=a\times 5^3\Rightarrow 125=125a\Rightarrow a=\frac{125}{125}=1$

Now, with $a=1$ and $r=5$, the required terms of G.P are:

$a_1=1a_2=a_1\times r=1\times 5=5a_3=a_2\times r=5\times 5=25a_4=a_3\times r=25\times 5=125$

Hence the G.P is $1,5,25,125,.....$.