Question

If in a G.P. r = 2 and t₈ = 64 then find a and S₆.

Answer 1

We know that the n^th term of the G.P. is t_n = ar^{n – 1}.
We know: r = 2 and t₈ = 64
Thus, we have:
t₈ = a(2)^{8 – 1}
$\Rightarrow$a(2)⁷ = 64
$\Rightarrow$a × 128 = 64
$\Rightarrow$a =$\frac{64}{128}=\frac{1}{2}$
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.
To find S₆, we will take a =$\frac{1}{2}$, r = 2 and n = 6.
Thus, we have:

$$\begin{gathered} S_6=\frac{{\displaystyle \frac{1}{2}}\left({\left(2\right)}^6-1\right)}{2-1} \\ =\frac{64-1}{2\times 1} \\ =\frac{63}{2} \end{gathered}$$