Question

$\huge\bold\red{kaho Kaisan ba solve kr da na piliz:-}$
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Answer 1

$\sf \purple{ \large\underline{Question}}$

• find the sum of a geometric series in which a = 1 /64, r=4, Tn = 16

$\sf \pink{ \large\underline{given}}$

$\sf \: \begin{lgathered} \sf \: a = \frac{1}{64} \:\: \\ \sf \: r = 4 \\\end{lgathered} \\ </p><p> \sf \text{ \red{on \: applying \: }} \\ </p><p> \sf \: \begin{lgathered} \sf \: T_n = 16 \\ \\ \sf \to a {r}^{n - 1} = 16 \\ \\ \sf \to \frac{1}{64} \times {4}^{n - 1} = 16 \\ \\ \sf \to \: {4}^{n - 1} = 16 \times 64 \\ \\ \sf \to {4}^{n - 1} = {4}^{2} \times {4}^{3} \\ \\ \sf \to {4}^{n - 1} = {4}^{5} \\\end{lgathered}$

$\sf \text{ \green{On comparing power of 4}}$

$</p><p> \begin{lgathered} \sf \: n - 1 = 5 \\ \\ \sf n = 6 \\\end{lgathered} </p><p>$

$\sf \text{\orange{Sum of given geometric series}}$

$\begin{lgathered} \sf \red{S_n = \frac{a( {r}^{n } - 1) }{r - 1} } \\ \sf \text{ \pink{putting \: all \: values}}\\ \\ \sf \to \frac{ \frac{1}{64}( {4}^{6} - 1) }{(4 - 1)} \\ \\ \sf \to \frac{4096 - 1}{64 \times 3} \\ \\ \sf \to \frac{4095}{64 \times 3} \\ \\ \sf \to \red{ \frac{1365}{64}}\end{lgathered} </p><p>$

Hence, Sum of given geometric series is 1365/64