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If the Sum of infinite terms in a GP is 12 and the first term is 8,
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11 Mar 2022, 23:18
11 Mar 2022, 23:18
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If the Sum of infinite terms in a GP is 12 and the first term is 8, then what is the 4th term of this GP?
A. \(\frac{8}{27}\)
B. \(\frac{4}{27}\)
C. \(\frac{8}{20}\)
D. \(\frac{4}{20}\)
E. \(\frac{1}{3}\)
A. \(\frac{8}{27}\)
B. \(\frac{4}{27}\)
C. \(\frac{8}{20}\)
D. \(\frac{4}{20}\)
E. \(\frac{1}{3}\)
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
If the Sum of infinite terms in a GP is 12 and the first term is 8,
[#permalink]
25 Mar 2022, 01:23
25 Mar 2022, 01:23
1
OA Explanation:
Sum of infinite terms in a GP is given be, \(S_∞ = \frac{a}{(1 - r)}\), where a is the first term and r is the common ratio (r < 1)
\(12 = \frac{8}{(1 - r)}\)
\((1 - r) = \frac{8}{12} = \frac{2}{3}\)
i.e. \(r = \frac{1}{3}\)
nth term in GP is given by, \(T_n = ar^{n - 1}\)
\(T_4 = 8(\frac{1}{3})^{4 - 1}\)
\(T_4 = 8(\frac{1}{3})^3 = (\frac{8}{27})\)
Hence, option A
Sum of infinite terms in a GP is given be, \(S_∞ = \frac{a}{(1 - r)}\), where a is the first term and r is the common ratio (r < 1)
\(12 = \frac{8}{(1 - r)}\)
\((1 - r) = \frac{8}{12} = \frac{2}{3}\)
i.e. \(r = \frac{1}{3}\)
nth term in GP is given by, \(T_n = ar^{n - 1}\)
\(T_4 = 8(\frac{1}{3})^{4 - 1}\)
\(T_4 = 8(\frac{1}{3})^3 = (\frac{8}{27})\)
Hence, option A
gmatclubot
If the Sum of infinite terms in a GP is 12 and the first term is 8, [#permalink]
25 Mar 2022, 01:23
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