Q17 of 176 Page 8

How many terms of the A.P. 3, 5, 7, 9, ... must be added to get the sum 120?

AP = 3, 5, 7, 9, …


Here, a = 3, d = 5 – 3 = 2 and Sn = 120


We know that,







⇒ 120 = n[2+n]


⇒ n2 + 2n – 120 = 0


⇒ n2 + 12n – 10n – 120 = 0


⇒ n(n + 12) - 10(n + 12) = 0


⇒ (n – 10)(n + 12) = 0


⇒ n – 10 = 0 or n + 12 = 0


⇒ n = 10 or n = -12


But number of terms can’t be negative. So, n = 10


Hence, for n = 10 the sum is 120 for the given AP.


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