Question
Given a G.P. with a = 729 and $7^{th}$ term 64, determine $S_7$ .
✓
Answer
Given: a = 729 and $a_7$ = 64$\Rightarrow a r ^ { 6 } = 64$
$\Rightarrow 729{r^6} = 64$
$\Rightarrow {r^6} = {{64} \over {729}} = {\left( {{2 \over 3}} \right)^6}$
$\Rightarrow r = \frac { 2 } { 3 }$
$\Rightarrow S _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r }$ when r < 1
$\Rightarrow S _ { 7 } = \frac { 729 \left[ 1 - \left( \frac { 2 } { 3 } \right) ^ { 7 } \right] } { 1 - \frac { 2 } { 3 } } = \frac { 729 \left[ 1 - \frac { 128 } { 2187 } \right] } { \frac { 3 - 2 } { 3 } }$
$\Rightarrow S _ { 7 } = 729 \times 3 \left( \frac { 2187 - 128 } { 2187 } \right)$
$\Rightarrow \mathrm { S } _ { 7 } = \frac { 729 \times 3 \times 2059 } { 2187 } = 2059$
$\Rightarrow 729{r^6} = 64$
$\Rightarrow {r^6} = {{64} \over {729}} = {\left( {{2 \over 3}} \right)^6}$
$\Rightarrow r = \frac { 2 } { 3 }$
$\Rightarrow S _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r }$ when r < 1
$\Rightarrow S _ { 7 } = \frac { 729 \left[ 1 - \left( \frac { 2 } { 3 } \right) ^ { 7 } \right] } { 1 - \frac { 2 } { 3 } } = \frac { 729 \left[ 1 - \frac { 128 } { 2187 } \right] } { \frac { 3 - 2 } { 3 } }$
$\Rightarrow S _ { 7 } = 729 \times 3 \left( \frac { 2187 - 128 } { 2187 } \right)$
$\Rightarrow \mathrm { S } _ { 7 } = \frac { 729 \times 3 \times 2059 } { 2187 } = 2059$
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