Question on: JAMB Mathematics - 2011
The seconds term of a geometric series is 4 while the fourth term is 16. Find the sum of the first five terms
A
60
B
62
C
54
D
64
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Correct Option: B
T2 = 4, T4 = 16
Tx = arn-1
T2 = ar2-1 = 4 i.e. ar3 = 16, i.e. ar = 4
T4 = ar4-1
therefore, \(\frac{T_4}{T_r}\) = \(\frac{ar^3}{ar}\) = \(\frac{16}{4}\)
r2 = 4 and r = 2
but ar = 4
a = \(\frac{4}{r}\) = \(\frac{4}{2}\)
a = 2
Sn = \(\frac{a(r^n - 1)}{r - 1}\)
S5 = \(\frac{2(2^5 - 1)}{2 - 1}\)
= \(\frac{2(32 - 1)}{2 - 1}\)
= 2(31)
= 62
Tx = arn-1
T2 = ar2-1 = 4 i.e. ar3 = 16, i.e. ar = 4
T4 = ar4-1
therefore, \(\frac{T_4}{T_r}\) = \(\frac{ar^3}{ar}\) = \(\frac{16}{4}\)
r2 = 4 and r = 2
but ar = 4
a = \(\frac{4}{r}\) = \(\frac{4}{2}\)
a = 2
Sn = \(\frac{a(r^n - 1)}{r - 1}\)
S5 = \(\frac{2(2^5 - 1)}{2 - 1}\)
= \(\frac{2(32 - 1)}{2 - 1}\)
= 2(31)
= 62
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