Q4 of 202 Page 9

If the n sum of a certain number of terms starting from first term of an A.P. is is 116. Find the last term.

a = 25, d = -3

Sn = [2a + (n – 1) d]


116 = [50 – 3n +3]


116 = [53 – 3n]


232 = 53n – 3n2


3n2 – 53n + 232 = 0


n = 8 or n = , which isn’t possible as n must be a natural number.


Therefore, n = 8


[a + l] = 116


[25 + l] = 116



l = 4


Hence, the last term is 4


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