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06 Dec 2022, 09:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
52% (02:42) correct
48%
(02:47)
wrong
based on 193
sessions
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Arithmetic sequences \(S_1\) and \(S_2\) have 5 terms each. If the first terms of \(S_1\) and \(S_2\) are \(a_1\) and \(a_2\) respectively, and the difference between two consecutive terms of \(S_2\) is twice the difference between two consecutive terms of \(S_1\), what is the ratio of the fifth term of \(S_1\)and the fifth term of \(S_2\)?
(1) \(a_1= 2a_2\)
(2) The second term of both \(S_1\) and \(S_2\) is 3
(1) \(a_1= 2a_2\)
(2) The second term of both \(S_1\) and \(S_2\) is 3
16 Dec 2022, 11:41
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Bunuel
5th term of \( s_1 : a_1 + 4x\)
5th term of \(s_2 :a_2 + 4 (2x)\)
We need : \(\frac{ a_1 + 4x }{a_2 + 4 (2x)}\) ... (I)
(1) \(a_1= 2a_2\)
\(\frac{ 2a_2 + 4x }{a_2 + 4 (2x)}\)
\(\frac{2( a_2 + 2x )}{a_2 + 8x}\)
INSUFF.
(2) The second term of both \(S_1\) and \(S_2\) is 3
\(a_1 + x = 3\) ...(II)
\(a_2 + 2x = 3\) ...(III)
\(a_1 - a_2 = x\) ... ( II - III )
\(a_1 = x+ a_2\) ... (IV)
Putting the value of \(a_1\) in ...(I)
\(\frac{ a_2+ 5x }{a_2 + 8x}\)
INSUFF.
1+2
From (1) We know \(a_1 = 2a_2 \)
From (2) We know \(a_1 = x+a_2 \\
\)
So \(2a_2 = x+a_2 \)
\(a_2 = x \)
Therefore \( a_1 = 2x \)
Putting all value in (I)
\(\frac{ 2x + 4x }{x + 8x}\)
\(\frac{6x}{9x} = \frac{6}{9}\)
SUFF.
Ans C
Hope it helped.
04 Apr 2026, 07:50
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Bunuel, isn't B sufficient?
Assuming d1 to be the common difference for S1 and d2 to be the common difference for S2, and as given in the question d2=2d1
We know S1= a1, a1+d1, a1+2d1....a1+4d1
S2= a2, a2+d2, a2+2d2..a2+4d2
As per B,
a1+d1=3 -> (1)
and a2+d2=3 -> (2)
hence a1+d1=a2+d2
We know this means a1+d1=a2+2d1, rephrased as a1-a2=d1.
Now from (1) & (2),
3-d1+3-d2=d1, rephrased as 3-d1+3-2d1=d1.
Therefore, 6=4d1 or d1=1.5
Substitute in (1) & (2), and you get a1=3-1.5=1.5 and a2=3-3=0
Hence a1=1.5, d1=1.5 and a2=0, d2=3
The question asks us for the value of \(\frac{a1+4d1}{a2+4d2}\)
We substitute the values above to get a complete answer i.e \(\frac{7.5}{12}\)
Hence the statement is sufficient.
Please let me know if there is anything missing or done incorrectly.
Assuming d1 to be the common difference for S1 and d2 to be the common difference for S2, and as given in the question d2=2d1
We know S1= a1, a1+d1, a1+2d1....a1+4d1
S2= a2, a2+d2, a2+2d2..a2+4d2
As per B,
a1+d1=3 -> (1)
and a2+d2=3 -> (2)
hence a1+d1=a2+d2
We know this means a1+d1=a2+2d1, rephrased as a1-a2=d1.
Now from (1) & (2),
3-d1+3-d2=d1, rephrased as 3-d1+3-2d1=d1.
Therefore, 6=4d1 or d1=1.5
Substitute in (1) & (2), and you get a1=3-1.5=1.5 and a2=3-3=0
Hence a1=1.5, d1=1.5 and a2=0, d2=3
The question asks us for the value of \(\frac{a1+4d1}{a2+4d2}\)
We substitute the values above to get a complete answer i.e \(\frac{7.5}{12}\)
Hence the statement is sufficient.
Please let me know if there is anything missing or done incorrectly.
