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Mathematics-2017-->View question

By:Purnima

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## If, for a positive integer n, the quadratic equation, x(x + 1) + (x + 1) (x + 2) + .....+ (x + n-1 ) (x + n) = 10n has two consecutive integral solutions, then n is equal to :-JEE MAINS-Mathematics-2017

(1) 11 (2) 12

(3) 9 (4) 10

By:Purnima

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## Answers

We have

(x+r-1)(x+r)=10n

(x^2+(2r-1)x+(r^2-r))=10n

On solving,
we get

x^2 + nx + ((n^2-31)/3)=0

/ \

a a+1

\ (2a + 1) = –n

a = -(n + 1)/2 ...(1)

and a(a +1) = (n^2-31)/3 ...(2)

n^2 = 121
(using (1) in (2))

thus,n = 11

Ans)(a)

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