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21 Feb 2022, 22:45
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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:
75%
(hard)
Question Stats:
56% (02:09) correct
44%
(01:56)
wrong
based on 189
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Let \(s_1\) be the sum of the first n terms of the arithmetic sequence 8, 12, . . . and let \(s_2\) be the sum of the first n terms of the arithmetic sequence 17, 19 . . . Assume n\(\neq\) 0. Then \(s_1\) = \(s_2\) for:
A. no value of n
B. one value of n
C. two values of n
D. four values of n
E. more than four values of n
A. no value of n
B. one value of n
C. two values of n
D. four values of n
E. more than four values of n
21 Feb 2022, 23:21
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twobagels
\(s_1\) is an AP with common difference of 4. => 8, 12, . . . (8+4n)
\(s_2\) is an AP with common difference of 2. => 17, 19 . . . (17+2n)
Since the number of terms is SAME, the SUMs will be same when the average is same => \(\frac{8+(8+4n)}{2}=\frac{17+(17+2n)}{2}........16+4n=34+2n....n=9\)
Thus, only one value 9.
B
07 Jul 2023, 09:20
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twobagels
Solution:
- First AP \(= 8, 12, . . . \)
- Sum of n terms \(s_1=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}[2\times 8+(n-1)4]=\frac{n}{2}[16+4n-4]=\frac{n}{2}[12+4n]\)
- Second AP \(= 17, 19 . . .\)
- Sum of n terms \(s_2=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}[2\times 17+(n-1)2]=\frac{n}{2}[34+2n-2]=\frac{n}{2}[32+2n]\)
- We are given \(s_1=s_2\)
\(⇒\frac{n}{2}[12+4n]=\frac{n}{2}[32+2n]\)
\(⇒12+4n=32+2n\) (Note that we are able to cancel \(n/\)2 from both sides because of the inference that \(n≠0\))
\(⇒2n=20\)
\(⇒n=10\)
Hence the right answer is Option B
