How many times of AP 24,21,18.....must be taken such that there sum is

1.	How many times of AP 24,21,18.....must be taken such that there sum is 78
Answer» sorry in above qn on AP please note the correction as n=13 , 4 in place of n=14, 3 thank u let number of terms = na = 24d = -378 = n/2 {2a + (n - 1) (-3)78 = 24n - 3n2/2 + 3n/2156 = 48n - 3n2 + 3n3n2 - 51n + 156 = 0n2 -17n + 52 = 0n2 - 14n - 3n + 52 = 0n(n - 14) - 3 (n - 14) =0n = 14 or n = 3\xa0

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