The sum of three consecutive terms of an AP is 21 and the product of the first and the third terms exceed tge second term by 6

JATIN SHARMA's image
JATIN SHARMA
Feb. 8, 2019, 11:45 p.m.

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Andleeb Mahtab's image
Andleeb Mahtab
Feb. 9, 2019, 11:23 a.m.

Let the 3 terms in AP be a-d,a and a+d

then a-d+a+a+d=21

3a=21, a=7

now, (a-d)(a+d)=a+6

ie, (7-d)(7+d)=13

d^2=36

d=-6 or 6

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