Question

The 12th term of an AP is - 13 and the sum of its first four terms is 24. Find the sum of its first 10 terms.

Answer 1

Let a and d are the first term and the common difference of an A P.

nth term = $t_n$ = a+ ( n - 1 ) d
Given; $t_{12}$ = - 13
a + 11d = -13 ---------(i)

As per question; sum of first 4 terms = 24
$S_n$= $\frac{n}{2}$ ($2a+(n-1)d$)
$S_4$ = 24
2 [ 2a + 3d] = 24
2[2a + 3d ] = 24
2a + 3d = 12 ------------(ii )

Multiply equation ( i ) with 2 and subtract it from (ii)
We get;
-19 d = 38
$\Rightarrow$d = -2
Substituting the value of d in (i)
a - 22 = -13
$\Rightarrow$a = -13 + 22
$\Rightarrow$ a = 9

We have:

a = 9 , d = -2 , n =10
$S_{10}$ = 5 [ 2 × 9 + ( 10 - 1 ) ( - 2 ) ]
= 5 [ 18 - 18 ]
= 5 × 0
= 0
Therefore, sum of first 10 terms = $S_{10}$ = 0