Question

the 12th term of an ap is minus 13 and the sum of its first 4 terms is 24 find the sum of the first 10 terms

Answer 1

T12 = - 13
a+ 11d = - 13
2a + 22d = - 26 --------(1)

T1 + T2 + T3 + T4 =24
=> a +a +d +a +2d +a +3d = 24
=> 4a + 6d = 24
=> 2a +3d = 12 ----------(2)

On subtracting equation 2 from equation 1, we get

19d = - 38
=> d = -2

a = 9


S10 = 10[2*9 +(10-1)(-2)]/2
= 5 (18 - 18)
= 5 * 0
= 0
S10 = 0