Q32 of 202 Page 10

If , then n =


Now, Sum of n terms


⇒


=


= n[5 + 2n–2]


=n[3 + 2n]


Now,


⇒


=


= (n + 1)[7 + n]



⇒


On cross multiplying we get,


16n[3 + 2n] = 17n + 17[7 + n]


⇒ 48n + 32n2 = 119n + 17n2 + 119 + 17n


⇒ 48n + 32n2 = 136n + 17n2 + 119


⇒ 15n2 – 88n – 199 = 0


⇒ n =

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