Question

Find the sum of the sequence $96 - 48 + 24 - .....$ up to 10 terms.

Answer 1

Hint – In the series given the common ratio that is the ratio of consecutive numbers remains constant throughout, thus it is forming an G.P. Use the direct formula for sum of n terms of an G.P which is ${S_n} = \dfrac{{{a_1}\left( {1 - {r^n}} \right)}}{{1 - r}}$.
Complete Step-by-Step solution:
Given series
$96 - 48 + 24 - .....$
We have to find out the sum of this series up to 10 terms.
Now as we see that the first term (a1) of the series is 96.
And the common ratio (r) = $\left( {\dfrac{{ - 48}}{{96}}} \right) = \left( {\dfrac{{24}}{{ - 48}}} \right) = - \dfrac{1}{2}$
As we see that the common ratio is equal so this series forms a G.P.
So here we use the sum (Sn) of G.P which is given as when common ratio (r) < 1,
$ \Rightarrow {S_n} = \dfrac{{{a_1}\left( {1 - {r^n}} \right)}}{{1 - r}}$ Where symbols have their usual meanings.
Now substitute the values in above equation we have,
$ \Rightarrow {S_n} = \dfrac{{96\left( {1 - {{\left( {\dfrac{{ - 1}}{2}} \right)}^{10}}} \right)}}{{1 - \left( {\dfrac{{ - 1}}{2}} \right)}}$
Now simplify the above equation we have,
$ \Rightarrow {S_n} = \dfrac{{96\left( {1 - \dfrac{1}{{1024}}} \right)}}{{1 + \dfrac{1}{2}}} = \dfrac{{96\left( {\dfrac{{1023}}{{1024}}} \right)}}{{\dfrac{3}{2}}} = \dfrac{{96\left( {1023} \right)\left( 2 \right)}}{{1024\left( 3 \right)}} = 63.9375$
So this is the required sum of the given series up to 10 terms.
So this is the required answer.

Note – The trick point here was that the common ratio that is $r = - \dfrac{1}{2}$ is less than 1 that’s why we have used the formula ${S_n} = \dfrac{{{a_1}\left( {1 - {r^n}} \right)}}{{1 - r}}$, if the common ratio would have been greater than 1 than it that case a different formula for the sum of n terms in G.P is used which is ${S_n} = \dfrac{{{a_1}\left( {{r^n} - 1} \right)}}{{r - 1}}$.