Q. 263.9( 17 Votes )

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 :

Let a be the first term and d be the common difference.

Given: a12 = - 13


S4 = 24


To find: Sum of first 10 terms.


Consider a12 = - 13


a + 11d = - 13 ………………(1)


Also, S4 = 24


(4/2) × [2a + (4 - 1)d] = 24


2 × [2a + 3d] = 24


2a + 3d = 12 …………….(2)


Subtracting equation (2) from twice of equation (1), we get,


19d = - 38


d = - 2


Now, from equation (1), we get


a = - 13 - 11d


a = - 13 - 11(-2)


a = - 13 + 22


a = 9


Now, Sum of first n terms of this arithmetic series is given by:


Sn = [2a + (n - 1)d]


Therefore sum of first 10 terms of this arithmetic series is given by:


S10 = [2(9) + (10 - 1)(-2)]


= 5 × [18 - 18]


= 0


S10 = 0


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