how many terms are there in an AP whose first term is -14, common difference is 4 and sum of terms is 40?
Sn = n/2 [2a +(n-1)d]
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.
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.