Question

The product of two consecutive terms of arithmetic sequence 5,8,11... is 598. Find the positions of the terms multiplied ?

Answer 1

Answer: here , a = 5 , d = 3

Let the position of the term be n+1and n+2 .

Therefore

( a+ nd )(a + (n+ 1)d) = 598

Simplifying this equation we get ,

9n^2 + 39n - 558 = 0

3n^2 + 13n - 186 = 0

3n^2 + 31n - 18 n - 186 = 0

n(3n + 31 ) - 6(3n + 31 ) = 0

(n - 6)(3n + 31) = 0

Therefore , n = 6 or n = -31/3

Here n = -31/3 is neglected .

The positions of the terms multiplied are 7 and 8 .

Hope it helps

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