If 12th term of an A.P. is $-13$ and the sum of the first four terms is 24, what is the sum of first 10 terms?
Given:
The 12th term of an A.P. is $-13$ and the sum of the first four terms is 24.
To do:
We have to find the sum of the first 10 terms.
Solution:
Let the first term of the A.P. be $a$ and the common difference be $d$.
We know that,
Sum of $n$ terms $S_n=\frac{n}{2}(2a+(n-1)d)$
$S_4=\frac{4}{2}[2a+(4-1)d]$
$24=2(2a+3d)$
$2a+3d=12$
$2a=12-3d$
$a=\frac{12-3d}{2}$......(i)
$a_{12}=a+(12-1)d$
$-13=a+11d$
$-13=\frac{12-3d}{2}+11d$
$-13=\frac{12-3d+2(11d)}{2}$
$-13(2)=12-3d+22d$
$-26=12+19d$
$19d=-26-12$
$d=\frac{-38}{19}$
$d=-2$
This implies,
$a=\frac{12-3(-2)}{2}$
$=\frac{12+6}{2}$
$=\frac{18}{2}$
$=9$
Therefore,
$S_{10}=\frac{10}{2}[2(9)+(10-1)(-2)]$
$=5(18-18)$
$=0$
Hence, the sum of the first 10 terms is $0$.
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