Question

# In an A.P if the $12^{th}$ term is -13 and the sum of its first four terms is 24. Find the sum of its first ten terms.

Hint: Assume an A.P series with ‘a’ as its first term and the difference between the consecutive terms be ‘d’. Our $12^{th}$ term of the A.P is -13. So, we can write it in mathematical form, i.e, $a+11d=-13$ . We have a summation of the first four terms equal to 24. So, we can also write it in mathematical form, i.e, $4a+6d=24$. Now, we have two equations and two unknown variables. Using these two equations, find the values of ‘a’ and ‘d’. Now, it can be solved further and we can find the summation of the first ten terms.

Complete step-by-step solution -
Let us assume an A.P with ‘a’ as its first term and the difference between the consecutive term be ‘d’.
We have the value of the $12^{th}$ term. So, first of all, we have to find the $12^{th}$ term in terms of the variable ‘a’ and ‘d’.
Putting n=12 in the equation, ${{T}_{n}}=a+(n-1)d$ , we get
${{T}_{12}}=a+(12-1)d$ ……………….(1)
According to the question, we have the $12^{th}$ term of an A.P as -13.
Now, we can write equation (1), as
$a+11d=-13$ ……………(2)
We know that the summation of n terms of an A.P is ${{S}_{n}}=\dfrac{n}{2}\left( {{1}^{st\,term}}+\operatorname{Last}\,term \right)$ .
According to the question, we have the summation of the first four terms which is 24.
${{S}_{n}}=\dfrac{n}{2}\left( {{1}^{st\,term}}+\operatorname{Last}\,term \right)$
$24=\dfrac{4}{2}\left( {{1}^{st\,term}}+\operatorname{Fourth}\,term \right)$ ……………(3)
Now, we have to find the $4^{th}$ term of the A.P.
Putting n=4 in the equation, ${{T}_{n}}=a+(n-1)d$ , we get
${{T}_{4}}=a+(4-1)d=a+3d$ …………….(4)
Now, putting equation (4) in equation (3) and we have ‘a’ as the first term of the A.P.
$24=2(a+a+3d)$
$\Rightarrow 12=2a+3d$ ……………………(5)
Using equation (2) and equation (5), we can find the value of ‘a’ and ‘d’.
According to equation (2), we have
$a+11d=-13$
$\Rightarrow a=-11d-13$ …………..(6)
Putting the value of ‘a’ from equation (6) in equation (5), we get
\begin{align} & 12=2a+3d \\ & \Rightarrow 12=2(-11d-13)+3d \\ \end{align}
\begin{align} & \Rightarrow 12=-22d-26+3d \\ & \Rightarrow 12+26=-22d+3d \\ & \Rightarrow 38=-19d \\ & \Rightarrow -2=d \\ \end{align}
We have got the value of ‘d’ which is -2.
Now, putting the value of ‘d’ in equation (6), we get
\begin{align} & a=-11d-13 \\ & \Rightarrow a=-11(-2)-13 \\ & \Rightarrow a=22-13 \\ & \Rightarrow a=9 \\ \end{align}
The 1st term of the A.P is 9.
Now, we have to find the summation of the first ten terms of the A.P.
Putting n=10 in the equation, ${{S}_{n}}=\dfrac{n}{2}\left( {{1}^{st\,term}}+\operatorname{Last}\,term \right)$ , we get
${{S}_{10}}=\dfrac{10}{2}\left( {{1}^{st\,term}}+\operatorname{Last}\,term \right)$ ………….(7)
Our last term is the $10^{th}$ term of the A.P. So, we have to find the $10^{th}$ term of the A.P.
Putting n=10 in the equation, ${{T}_{n}}=a+(n-1)d$ , we get
${{T}_{10}}=a+(10-1)d$
$\Rightarrow {{T}_{10}}=a+9d$ …………………(8)
Now, putting the values of ‘a’ and ‘d’ in equation (8), we get
${{T}_{10}}=9+9(-2)=-9$ …………..(9)
Now, we have got the $10^{th}$ term of the A.P.
Now, putting equation (9) in equation (7), we get
\begin{align} & {{S}_{10}}=\dfrac{10}{2}\left( {{1}^{st\,term}}+\operatorname{Last}\,term \right) \\ & \Rightarrow {{S}_{10}}=5(9-9)=0 \\ \end{align}
So, the summation of the first ten-term of the A.P is 0.

Note: We can also solve the summation using the formula, ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$ .
We have, n=10, a=9 and d=-2.
Putting the values of a,n and d in the above formula, we get
${{S}_{10}}=\dfrac{10}{2}\left[ 2(9)+(10-1)(-2) \right]=5\left[ 18-18 \right]=0$ .
Hence, the summation of the first ten terms of the A.P is 0.