# how many terms are there in an AP whose first term is -14, common difference is 4 and sum of terms is 40?

40 = n/2[2x(-14) + (n-1)4]

40 = n/2 [-28 + 4n-4]

40 = n/2 (4n-32)

40 = 2n sq. -16n

2n sq. - 16n - 40 = 0

n sq. - 8n -20 = 0

n sq. - 10n + 2n -20 = 0

n(n-10) + 2(n-10) = 0

n = 10

thus, there are 10 terms in the given AP.

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