Question

13. If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the

last term​

Answer 1

Answer :

4

Solution :

Given,

AP : 25, 22, 19, ....

first term, a = 25

common difference, d = 22 - 25 = -3

Sum of n terms = 116

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

116 = n/2 [50 + (n - 1)-3]

232 = 50n - 3n² + 3n

3n² - 53n + 232 = 0

3n² − (29+24)n+232=0

⇒3n² − 29n − 24n + 232=0

⇒n(3n−29 )− 8(3n−29)=0

⇒(3n−29)(n−8)=0

n = 8

The last term is

S_n = n/2[(a+l)]

116= 8/2 (25+l)

116=4(25+l)

116=100+4l

116−100=4l

4l=16

l=4

Hence, the last term of this series is

4.

This is the answer.