Question

The 14th term of an A.P is twice its 8th term. If its 6th term is -8, then find the sum of its first 20 terms.

Answer 1

Hint: Write 14th term, 8th term, 6th term of the A.P. with the help of the nth term for an A.P., given as
${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$

Complete step-by-step answer:
Where, ${{a}_{1}}$ is the first term, d is the common difference and ${{a}_{n}}$ is the nth term of an A.P.
Form two equations with the help of given two conditions in the problem. Use identity of sum of n terms to get the required answer, given
$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2{{a}_{1}}+\left( n-1 \right)d \right]$
n $\to$ number of terms
d $\to$ common difference
${{a}_{1}}\to$ first term

As we know Arithmetic progression is a sequence of numbers with same successive difference and general term or nth term of A.P. is given
$\Rightarrow {{a}_{n}}={{a}_{1}}+\left( n-1 \right)d...............\left( i \right)$
Where, ${{a}_{n}}$ is the nth term
${{a}_{1}}$ = First term of A.P.
d = common difference
As, we also know that sum of the terms up to n terms of an A.P. is also given
$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2{{a}_{1}}+\left( n-1 \right)d \right].............\left( ii \right)$
${{S}_{n}}$ = Sum up to n terms.
${{a}_{1}}$ = First term
d = common difference
Now, coming to the question, we are given that the 14th term of an A.P. is twice of the 8th term and 6th term of the same A.P. is -8 and hence, we have to determine the sum of the A.P. up to 20 terms.
So, we can write equations
$\begin{aligned} & \Rightarrow {{a}_{14}}=2{{a}_{8}}............\left( iii \right) \\\ & \Rightarrow {{a}_{6}}=-8..............\left( iv \right) \\\ \end{aligned}$
Now, let us suppose the first term of the given A.P. is ${{a}_{1}}$ and the common difference is ‘d’. So, using equation (i) and (iii), we get

& \Rightarrow {{a}_{1}}+\left( 14-1 \right)d=2\left( {{a}_{1}}+\left( 8-1 \right)d \right) \\\ & \Rightarrow {{a}_{1}}+13d=2\left( {{a}_{1}}+7d \right) \\\ & \Rightarrow {{a}_{1}}+13d=2{{a}_{1}}+14d \\\ & \Rightarrow 13d-14d=2{{a}_{1}}-{{a}_{1}}={{a}_{1}} \\\ & \Rightarrow {{a}_{1}}=-d.................................\left( v \right) \\\ \end{aligned}$$ And using equation (iv) and (i), we get $\begin{aligned} & \Rightarrow {{a}_{1}}+\left( 6-1 \right)d=-8 \\\ & \Rightarrow {{a}_{1}}+5d=-8...................\left( vi \right) \\\ \end{aligned}$ Substituting ${{a}_{1}}=-d$ from equation (v) to the equation (vi), we get $\begin{aligned} & -d+5d=-8 \\\ & \Rightarrow 4d=-8 \\\ & \Rightarrow d=-2 \\\ \end{aligned}$ And, hence value of ${{a}_{1}}$ is given as ${{a}_{1}}=2$ Now, as we have to find sum of A.P. up to 20 terms, so using equation (ii), we get $\begin{aligned} & {{S}_{20}}=\dfrac{20}{2}\left[ 2\times 2+\left( 20-1 \right)\left( -2 \right) \right] \\\ & \Rightarrow {{S}_{20}}=10\left[ 4+19\times \left( -2 \right) \right] \\\ & \Rightarrow {{S}_{20}}=10\left[ 4-38 \right] \\\ & \Rightarrow {{S}_{20}}=10\times \left( -34 \right) \\\ & \Rightarrow {{S}_{20}}=-340 \\\ \end{aligned}$ Hence, the sum of given A.P. up to 20 terms is -340. Note: One may calculate ${{a}_{20}}$ and use the formula $S=\dfrac{n}{2}\left[ a+l \right]$ Where ‘a’ and ‘l’ are first and last terms of given A.P. and n is the number of terms involved. One may write all the 20 terms of the given A.P. and hence, he/she may add the terms individually. But it will be a long approach as here only 20 terms are involved, so one can get the answer but one can not use this approach for calculating sum for A.P with higher number of terms. So, remember the identities to solve these types of problems easily. And do not get confuse with the identities of other series i.e. G.P or H.P. So, be clear with the concepts and remember the formulae to calculate nth term and sum up to n terms of this series.