Question

If S₆ = 126 and S₃ = 14 then find a and r.

Answer 1

We know that the sum of the first n^th term of the G.P. is ${\mathrm{S}}_n=\frac{a(r^n-1)}{r-1}$ for r > 1.
We know:
S₃ = 14 and S₆ = 126
Thus, we have:

$$\begin{gathered} S_3=\frac{a(r^3-1)}{r-1}=14....\left(1\right) \\ {S}_6=\frac{a(r^6-1)}{r-1}=126....\left(2\right) \\ \mathrm{On}\mathrm{dividing}\mathrm{equation}\left(2\right)\mathrm{by}\left(1\right),\mathrm{we}\mathrm{get}: \\ \frac{a(r^6-1)}{r-1}\times \frac{r-1}{a(r^3-1)}=\frac{126}{14} \\ \Rightarrow \frac{{\left(r^3\right)}^2-1}{r^3-1}=9 \\ \Rightarrow r^3+1=9 \\ \Rightarrow r^3=8 \\ \Rightarrow r^3=(2{)}^3 \\ \Rightarrow r=2 \\ \mathrm{Now},\mathrm{on}\mathrm{putting}r=2\mathrm{in}(1),\mathrm{we}\mathrm{get}: \\ \frac{a\left(2^3-1\right)}{2-1}=14 \\ \Rightarrow \frac{a\times 7}{1}=14 \\ \Rightarrow a=\frac{14}{7}=2 \end{gathered}$$

Thus, we have:
a = 2 and r = 2