In a G.P. sum of n terms is 255, the last term is 128 and the common ratio is 2. Find n.

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Solution

Given $T_{n} = a r^{n - 1} = 128, r = 2$
$S_{n} = \frac{a (r^{n} - 1)}{r - 1} = 255 o r \frac{a r^{n - 1} . r - a}{r - 1} = 255 o r \frac{128.2 - a}{2 - 1} = 255 ∴ 256 - 255 = a o r a = 1$ .
Now $a r^{n - 1} = 128 ∴ 1. (2)^{n - 1} = 2^{7} ∴ n - 1 = 7 o r n = 8.$

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