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RenB
Joined: 13 Jul 2022
Last visit: 30 Apr 2026
Posts: 388
Given Kudos: 304
Location: India
Concentration: Finance, Nonprofit
Schools: Stanford '26 Wharton '26 (D) Booth '26 (D) Yale '26 (D) CBS '26 (A$$$) Darden '26 (WL) HBS '26 (D) Ross '26 (A$$$$) LBS '26 (D) Tuck '26 (A$$) Kellogg '26 (A)
GMAT Focus 1: 715 Q90 V84 DI82 
GPA: 3.74
WE:Corporate Finance (Consulting)
Schools: Stanford '26 Wharton '26 (D) Booth '26 (D) Yale '26 (D) CBS '26 (A$$$) Darden '26 (WL) HBS '26 (D) Ross '26 (A$$$$) LBS '26 (D) Tuck '26 (A$$) Kellogg '26 (A)
GMAT Focus 1: 715 Q90 V84 DI82
Posts: 388
10 Sep 2023, 02:43
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
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Question Stats:
69% (02:14) correct
31%
(02:31)
wrong
based on 291
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Two arithmetic sequences, S1 and S2, are such that the difference between any two consecutive terms of S2 is twice the difference between any two consecutive terms of S1. If the third term of S2 is twice the third term of S1, what is the ratio of the 5th term of Si to the 5th term of S2?
A 1:3
B 1:2
C 2:1
D 4:1
E Cannot be determined
A 1:3
B 1:2
C 2:1
D 4:1
E Cannot be determined
11 Sep 2023, 05:08
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Two arithmetic sequences, S1 and S2, are such that the difference between any two consecutive terms of S2 is twice the difference between any two consecutive terms of S1
Let common difference of Series S1 is d1 and
common difference of series S2 is d2
=> d2 = 2 * d1
If the third term of S2 is twice the third term of S1
Let first term of series S1 = a1 and
first term of series S2 = a2
Using Arithmetic series formula, \({T_3}\) = a + (3-1)*d = a + 2d
\({T_3}\) of S1 = a1 + 2*d1
\({T_3}\) of S2 = a2 + 2*d2 = a2 + 2*2*d1 = a2 + 4d1
\({T_3}\) of S1 = 2*\({T_3}\) of S1
=> a2 + 4d1 = 2 * (a1 + 2*d1) = 2a1 + 4d1
=> a2 = 2a1
what is the ratio of the 5th term of S1 to the 5th term of S2?
=> \(\frac{a1 + 4d1 }{ a2 + 4d2}\) = \(\frac{a1 + 4d1 }{ 2a1 + 4*2d1}\) = \(\frac{a1 + 4d1 }{ 2*(a1 + 4*d1)}\) = \(\frac{1}{2}\)
So, Answer will be B
Hope it helps!
Watch the following video to learn How to Sequence problems
Let common difference of Series S1 is d1 and
common difference of series S2 is d2
=> d2 = 2 * d1
If the third term of S2 is twice the third term of S1
Let first term of series S1 = a1 and
first term of series S2 = a2
Using Arithmetic series formula, \({T_3}\) = a + (3-1)*d = a + 2d
\({T_3}\) of S1 = a1 + 2*d1
\({T_3}\) of S2 = a2 + 2*d2 = a2 + 2*2*d1 = a2 + 4d1
\({T_3}\) of S1 = 2*\({T_3}\) of S1
=> a2 + 4d1 = 2 * (a1 + 2*d1) = 2a1 + 4d1
=> a2 = 2a1
what is the ratio of the 5th term of S1 to the 5th term of S2?
=> \(\frac{a1 + 4d1 }{ a2 + 4d2}\) = \(\frac{a1 + 4d1 }{ 2a1 + 4*2d1}\) = \(\frac{a1 + 4d1 }{ 2*(a1 + 4*d1)}\) = \(\frac{1}{2}\)
So, Answer will be B
Hope it helps!
Watch the following video to learn How to Sequence problems
11 Sep 2023, 17:32
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The most straightforward way I could think was to take random number for third term of s1 and double it for third number of s2
S1( 1,2,3,4,5)
Since the difference in any 2 numbers of s2 is twice that of s1 the third term will be 6
s2(2,4,6,8,10)
fifth term will be 5/10 =1/2
option B
S1( 1,2,3,4,5)
Since the difference in any 2 numbers of s2 is twice that of s1 the third term will be 6
s2(2,4,6,8,10)
fifth term will be 5/10 =1/2
option B
