Question

find the sum of he geometric series 3+6+12+..+1536

Answer 1

Solution :

$\\ \\ \\ \tt \longrightarrow \: 3 + 6 + 12 + ..+ 1536 \\ \\ \\ \tt \: here \: \: a \: = 3\\ \\ \\ \tt \longrightarrow \: r \: = \dfrac{6}{3} \\ \\ \\ \tt \longrightarrow \: r \: = \: 2\\ \\ \\ \tt \: now \: \: we \: \: have \: \: to \: \: find \: \: the \: \: value \: \: of \: \: n \\ \\ \\ \tt \: so \: \: we \: \: have \: \: a = 3 \: , \: r \: = 2\\ \\ \\ \tt \longrightarrow \: t_{n} \: = \: 1536\\ \\ \\ \tt \longrightarrow \: {ar}^{n - 1} \: = \: 1536\\ \\ \\ \tt \longrightarrow \:3 {(2)}^{n - 1} \: = 1536\\ \\ \\ \tt \longrightarrow \: {2}^{n - 1} \: = \: 512\\ \\ \\ \tt \longrightarrow \: {2}^{n - 1} \: = \: {2}^{9} \\ \\ \\ \tt \longrightarrow \: n - 1 \: = \: 9\\ \\ \\ \tt \longrightarrow \: n \: = \: 9 + 1\\ \\ \\ \tt \longrightarrow \: n = \: 10\\ \\ \\ \tt so \: \: now \: \: we \: \: have \: \: n = 10 \: , a = 3 \: , r = 2\\ \\ \\ \tt \: by \: \: using \: \: formula, \\ \\ \\ \tt \longrightarrow \: s_{n} \: = \dfrac{a( {r}^{n} - 1 )}{(r - 1)} \\ \\ \\ \tt \longrightarrow \: s_{n} \: = \: \dfrac{3( {2}^{10} - 1) }{2 - 1} \\ \\ \\ \tt \longrightarrow \: s_{n} \: = \: 3(1024 - 1)\\ \\ \\ \tt \longrightarrow \: s_{n} \: = \: 3 \times 1023\\ \\ \\ \tt \longrightarrow \: s_{n} \: = \: 3069 \: \: \:$

•°• Hence, The answer is 3069