Question

In an A.P if the 12th term is - 13 & the sum of its 1st four terms is 24, then find the sum of it's first ten terms

Answer 1

A12 = -13
a + 11d = -13
a = -13 - 11d -------(1)

Sum of first 4 terms
a + (a + d) + (a + 2d) + (a + 3d) = 24
4a + 6d = 24
2a + 3d = 12
substitute (1)
2 ( -13 -11d ) + 3d = 12
-26 -22d + 3d = 12
-19d = 38
d = -2

Putting in (1)
a = -13 - 11d
a = -13 - 11(-2)
a = -13 +22
a = 9

Sum of 10 terms,Sn=n/2(2a+(n-1)d) S10=10/2(2(9)+(10-1)(-2)) S10=5(18-18)
S10=0