Question

find the sum of a geometric series in which a = 1 /64, r=4, Tn = 16
(on the basis on the sum of number inGP)

Answer 1

Hi ,

Here is your answer !
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Given that,

$a = \frac{1}{64} \:, \: r = 4$
And

$T_n = 16 \\ \\ a {r}^{n - 1} = 16 \\ \\ \frac{1}{64} \times {4}^{n - 1} = 16 \\ \\ {4}^{n - 1} = 16 \times 64 \\ \\ {4}^{n - 1} = {4}^{2} \times {4}^{3} \\ \\ {4}^{n - 1} = {4}^{5}$
On comparing power of 4

$n - 1 = 5 \\ \\ n = 6$
Now,

Sum of given geometric series,

$S_n = \frac{a( {r}^{n } - 1) }{r - 1} \\ \\ = \frac{ \frac{1}{64}( {4}^{6} - 1) }{(4 - 1)} = \frac{4096 - 1}{64 \times 3} \\ \\ = \frac{4095}{64 \times 3} = \frac{1365}{64}$

Hence,

Sum of given geometric series is 1365/64

Answer 2

Answer:

solution of GP :-----

please Mark my answer brilliant

Step-by-step explanation: