Q. 3 F4.0( 22 Votes )

In an AP :

given an = 4, d = 2, sn = -14. Find n and a

Answer :

Given, series in A.P.

nth term in the AP an = 4,


Common difference (d) = 2


Sum of the series is sn = -14


an = a +(n-1)d


4 = a + (n-1)2


6 = a + 2n


a = 6 – 2n 1


Sum of the series (sn) = [2a + (n-1) d]


-14 = [ 2×(6-2n) + (n-1)2]


-14 = n[5-n]


n2 -5n -14 = 0


n2 -7n + 2n -14 = 0


n(n-7) + 2(n-7) = 0


(n-7)(n+2) = 0


n-7 = 0 n = 7 (or) n+2 = 0 n = -2


since n should be a positive integer


we take n = 7


substitute in equation 1


a = 6 – 2n


a = 6 – 2(7)


a = -8


first term is -8


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