Q. 32

# If , then n =A. 8B. 7C. 10D. 11 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 = Rate this question :

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