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04 Apr 2022, 06:15
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A certain sequence is defined by the following recursive rule: \(S_n = k*S_{n−1}\), where k is a constant. If the first term of the series is 4 and the 5th term is 324, what is the 7th term?
(A) 312
(B) 488
(C) 976
(D) 2916
(E) 6588
(A) 312
(B) 488
(C) 976
(D) 2916
(E) 6588
11 Apr 2022, 10:46
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\(S_1 = 4\)
\(S_2 = k*S_1 \)
\(S_3 = k*S_2 → k*k*S_1 → k^2*4\)
\(S_5 = 324 = k^4*4 → 81 = k^4 → k = 3\)
\(S_7 = k^6*4 → k^4*k^2*4 = 81*9*4\)
Since the answer choices are spaced part we can approximate \(81*9*4 \) to be little less than \(80*10*4 = 3200\)
Ans: D
\(S_2 = k*S_1 \)
\(S_3 = k*S_2 → k*k*S_1 → k^2*4\)
\(S_5 = 324 = k^4*4 → 81 = k^4 → k = 3\)
\(S_7 = k^6*4 → k^4*k^2*4 = 81*9*4\)
Since the answer choices are spaced part we can approximate \(81*9*4 \) to be little less than \(80*10*4 = 3200\)
Ans: D
31 Aug 2023, 07:03
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S1 = 4
S2 = k.S1
S3 = k^2. S1
Sn = k^n . S1
S5 = 324 = k^5. 4
k^5 = 81 implying k = 3
S7 = S5* 3*3 = 324*9 = number with units digit 6 and greater than 2700. So D.
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S2 = k.S1
S3 = k^2. S1
Sn = k^n . S1
S5 = 324 = k^5. 4
k^5 = 81 implying k = 3
S7 = S5* 3*3 = 324*9 = number with units digit 6 and greater than 2700. So D.
Posted from my mobile device
