Q37 of 176 Page 8

If the 12th term of an A.P. is - 13 and the sum of the first four terms is 24, what is the sum of the first 10 terms?

Given: a12 = -13


⇒ a + 11d = -13


⇒ a = -13 – 11d …(i)


and S4 = 24


[from(i)]


⇒ 2[-26 -22d + 3d] = 24


⇒ -26 – 19d = 12


⇒ -19d = 12 + 26


⇒ -19d = 38


⇒ d = -2


Putting the value of d in eq. (i), we get


a = -13 – 11(-2) = -13 + 22 = 9


So, a = 9 , d = -2 and n = 10


Now, we have to find the S10



⇒ S10 = 5[2×9 + 9(-2)]


⇒ S10 = 5[18 – 18]


⇒ S10 = 0


Hence, the sum of first 10 terms is 0


More from this chapter

All 176 →
35

If a, b, c be the 1st, 3rd and nth terms respectively of an A.P., there prove that the sum to n terms is

36

If the mth term of an A.P. is and the nth term is , then prove that the sum to mn terms is , where in m n.

 38

If the number of terms of an A.P. be 2n + 3, then find the ratio of sum of the odd terms to the sum of even terms.

 39

If the sum of first m terms of an A.P. is the same as the sum of its first n terms, show that the sum of its first (m + n) terms is zero.