Question

Find the sum of the following sequence $3, - 4,\dfrac{{16}}{3},.................{\text{to 2n}}$.

Answer 1

Look for the pattern of arithmetic progression or geometric progression in the given series. You will find out that this is a GP, find a common ratio and use the formula of summation of GP to get the answer.

Complete step-by-step answer:

The second term ${a_2} = - 4$ and first term ${a_1} = 3$, so the common ratio will be ${r_1} = \dfrac{{{a_2}}}{{{a_1}}}$.

On putting the values above we get our common ratio as $\dfrac{{ - 4}}{3}$…………………….. (1)

Now let’s verify that the common ratio is coming same by using the terms ${a_3}$ and ${a_2}$,

The third term ${a_3} = \dfrac{{16}}{3}$ and the second term is ${a_2} = - 4$ so the common ratio will be ${r_2} = \dfrac{{{a_3}}}{{{a_2}}}$.

So on putting the values above we get common ratio as $\dfrac{{\dfrac{{16}}{3}}}{{ - 4}} = \dfrac{{ - 4}}{3}$……………………… (2)

Now for a series to be in G.P the common ratio must be the same tha is ${r_1} = {r_2}$. Clearly from equation (1) and equation (2) we can say that

${r_1} = {r_2}$

Hence the given series is in G.P.

Now Sum of first $n$ terms of a G.P is given as

${s_n} = \dfrac{{a({r^n} - 1)}}{{r - 1}}$, Here $a$ is the first term r is the common ratio and n is the number of terms up to which sum is to be found that is $2n$.

Thus putting the values in sum formulae we get

${s_{2n}} = \dfrac{{3\left( {{{\left( {\dfrac{{ - 4}}{3}} \right)}^{2n}} - 1} \right)}}{{\dfrac{{ - 4}}{3} - 1}}$

On solving this we get

${s_{2n}} = \dfrac{{9\left( {{{\left( {\dfrac{{16}}{9}} \right)}^n} - 1} \right)}}{{ - 7}}$

Let’s take negative common from both numerator and denominator

${s_{2n}} = \dfrac{9}{7}\left[ {1 - {{\left( {\dfrac{4}{3}} \right)}^{2n}}} \right]$

Therefore, the sum of given series up to 2n terms is ${s_{2n}} = \dfrac{9}{7}\left[ {1 - {{\left( {\dfrac{4}{3}} \right)}^{2n}}} \right]$.

Note: Whenever we are given a series and told to find sum up to some terms always remember that the series will be an AP or will be a GP, even sometimes it can be HP. So simply using the basic series formulae for that particular category of series will help you reach the solution.