Question

The terms 1/64, 1/32, 1/16 form a geometric progression (GP) If the sum of the GP is (2^36 – 2^-6), find the number of terms

Answer 1

If $a$ is the first term and $r$ is the common ratio of the series, then the sum of the first $n$ terms is

$S_n = a + ar + ar^2 + \cdots + ar^n$

Multiply by $r$ on both sides, then subtract that from and solve for $S_n$.

$rS_n = ar + ar^2 + ar^3 + \cdots + ar^{n+1}$

$(1-r)S_n = a - ar^{n+1}$

$S_n = \dfrac{a(1-r^{n+1})}{1-r}$

Use the given first and second terms of the sequence to find $a$ and $r$.

$a = \dfrac1{64}$

$ar = \dfrac1{32} \implies \dfrac r{64} = \dfrac1{32} \implies r = \dfrac{64}{32}=2$

Solve for $n$.

$S_n = \dfrac{1-2^{n+1}}{64(1-2)} = 2^{36} - 2^{-6}$

$\dfrac{2^{n+1} - 1}{2^6} = 2^{36} - 2^{-6}$

$2^{n-5} - 2^{-6} = 2^{36} - 2^{-6}$

$\implies n-5 = 36 \implies \boxed{n = 41}$