Question

If S₃ = 31 and S₆ = 3906 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 = 31 and S₆ = 3906
Thus, we have:

$$\begin{gathered} S_3=\frac{a(r^3-1)}{r-1}=31.....\left(1\right) \\ {S}_6=\frac{a(r^6-1)}{r-1}=3906....\left(2\right) \\ \mathrm{On}\mathrm{dividing}\left(2\right)\mathrm{by}\left(1\right),\mathrm{we}\mathrm{get}: \\ \frac{a\left(r^6-1\right)}{r-1}\times \frac{r-1}{a\left(r^3-1\right)}=\frac{3906}{31} \\ \Rightarrow \frac{{\left(r^3\right)}^2-1}{r^3-1}=126 \\ \Rightarrow \frac{\left(r^3+1\right)\left(r^3-1\right)}{r^3-1}=126 \\ \Rightarrow r^3+1=126 \\ \Rightarrow r^3=125 \\ \Rightarrow r^3=(5{)}^3 \\ \Rightarrow r=5 \\ \mathrm{Now},\mathrm{on}\mathrm{putting}\mathit{}r=5\mathrm{in}(1),\mathrm{we}\mathrm{get}: \\ \frac{a\left(5^3-1\right)}{5-1}=31 \\ \frac{a\times 124}{4}=31 \\ a=\frac{31\times 4}{124}=1 \end{gathered}$$

Thus, we get:
a = 1 and r = 5