Question

Let the sum of n terms of an A.P be denoted by ${{S}_{n}}$. If ${{S}_{4}}=16$ and ${{S}_{6}}=-48$, then ${{S}_{10}}$ is equal to?
(A) -320
(B) -260
(C) -380
(D) -410

Answer 1

Hint: We start solving this problem by first using the formula for the sum of n terms of an A.P and find ${{S}_{n}}$ in terms of first term and common difference. Then we substitute the values $n=4$ and $n=6$ in ${{S}_{n}}$. Then we get two equations and we solve them to find the values of the first term and the common difference of A.P. Then let us substitute the value $n=10$ in ${{S}_{n}}$, and use the values of first term and the common difference to find the value of ${{S}_{10}}$.

Complete step-by-step answer:
We are given that the sum of n terms of an A.P is denoted by ${{S}_{n}}$.
Let the first term of the given A.P be $a$ and the common difference of given A.P be $d$. Then using the formula for sum of n terms of A.P, we get
${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$.
We are also given that ${{S}_{4}}=16$ and ${{S}_{6}}=-48$.
So, now let us substitute $n=4$ in ${{S}_{n}}$.
$\begin{align}
  & \Rightarrow {{S}_{4}}=\dfrac{4}{2}\left( 2a+\left( 4-1 \right)d \right) \\
 & \Rightarrow 16=2\left( 2a+3d \right) \\
 & \Rightarrow 2a+3d=8 \\
\end{align}$
So, now let us substitute $n=6$ in ${{S}_{n}}$.
$\begin{align}
  & \Rightarrow {{S}_{6}}=\dfrac{6}{2}\left( 2a+\left( 6-1 \right)d \right) \\
 & \Rightarrow -48=3\left( 2a+5d \right) \\
 & \Rightarrow 2a+5d=-16 \\
\end{align}$
So, now we have two equations with two variables.
$\begin{align}
  & \Rightarrow 2a+3d=8...........\left( 1 \right) \\
 & \Rightarrow 2a+5d=-16...........\left( 2 \right) \\
\end{align}$
Now, let us subtract equation (2) from equation (1). Then we get
$\begin{align}
  & \Rightarrow \left( 2a+5d \right)-\left( 2a+3d \right)=-16-8 \\
 & \Rightarrow 2d=-24 \\
 & \Rightarrow d=-12 \\
\end{align}$
Now let us substitute the obtained value of $d$ in equation (1).
$\begin{align}
  & \Rightarrow 2a+3\left( -12 \right)=8 \\
 & \Rightarrow 2a-36=8 \\
 & \Rightarrow 2a=36+8 \\
 & \Rightarrow 2a=44 \\
 & \Rightarrow a=22 \\
\end{align}$
So, we finally get the values as $a=22$ and $d=-12$.
As we need to find the value of ${{S}_{10}}$, let us substitute $n=10$ in the formula of ${{S}_{n}}$. Then we get,
$\begin{align}
  & \Rightarrow {{S}_{10}}=\dfrac{10}{2}\left( 2a+\left( 10-1 \right)d \right) \\
 & \Rightarrow {{S}_{10}}=5\left( 2\left( 22 \right)+9\left( -12 \right) \right) \\
 & \Rightarrow {{S}_{10}}=5\left( 44-108 \right) \\
 & \Rightarrow {{S}_{10}}=5\left( -64 \right) \\
 & \Rightarrow {{S}_{10}}=-320 \\
\end{align}$
So, we get the value of ${{S}_{10}}$ as -320.
Hence the answer is Option A.

Note: While solving this problem one might write the sum of 10 terms in A.P as ${{S}_{10}}={{S}_{4}}+{{S}_{6}}$. But it is wrong as in ${{S}_{4}}$ and ${{S}_{6}}$ first four terms are common and the last four terms are not included in the sum.