The first, twelfth and the last term of an arithmetic progression are 4, 31.5 and 376.5 respectively. Find the number of terms in the series?
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Step-by-step explanation:
In the AP, a = 37, Sn = 114 and Tn = 18, we want to know n.
Sn = 114 = (n/2)[2a+(n-1)d] …(1)
Tn = 18 = a+(n-1)d = 37 + (n-1)d, or
(n-1)d = 18–37 = -19 …(2)
Substitute (n-1)d from (2) in (1)
114 = (n/2)[74–19] = (n/2)*55
n = 114*2/55 = 4.14.
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