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In a sequence of integers, each term in the first m integers is 5 greater than the previous integer. Thereafter, each integer is 2 less than the previous integer. If the first integer and the last integer in the sequence are each 50 and the number of integers in the sequence is 29, m=?
A. 7
B. 9
C. 10
D. 12
E. 15
Source: Experts Global
A. 7
B. 9
C. 10
D. 12
E. 15
Source: Experts Global
24 Apr 2018, 00:46
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pushpitkc
In a sequence of integers, each term in the first m integers is 5 greater than the previous integer. Thereafter, each integer is 2 less than the previous integer. If the first integer and the last integer in the sequence are each 50 and the number of integers in the sequence is 29, m=?
A. 7
B. 9
C. 10
D. 12
E. 15
Source: Experts Global
A. 7
B. 9
C. 10
D. 12
E. 15
Source: Experts Global
First, we have m integers, for which each term is 5 greater than the previous integer:
50, 55, 60, 65, 70, ..., 50 + 5(m - 1)
Notice, that the last integer in this, first, part of the sequence is \(50 + 5(m - 1)\). \(n_{th}\) term in the arithmetic progression (AP) is given by \(a_n=a_1+d(n-1)\), where d is the common difference.
After those m integers, each integer is 2 less than the previous integer:
\([50 + 5(m - 1)] - 2\), \([50 + 5(m - 1)] - 4\), \([50 + 5(m - 1)] - 6\), ..., 50.
Notice, that since there are a total of 29 integers in the whole sequence, then the above continues for 29 - m integers. The last integer given to be 50, thus by applying the same rule for AP, we'll get \(50 = [50 + 5(m - 1) - 2] + (-2)[(29 - m) - 1]\):
\(0 = 5m - 7 -56 + 2m\)
\(7m=63\);
\(m=9\).
Answer: B.
12. Sequences
- Theory
Sequences Made Easy - All in One Topic!
Math : Sequences & Progressions
Time to Tackle Geometric Progressions!
For other subjects check:
ALL YOU NEED FOR QUANT.
Ultimate GMAT Quantitative Megathread
Hope it helps.
General Discussion
28 Mar 2018, 12:52
Kudos
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pushpitkc
In a sequence of integers, each term in the first m integers is 5 greater than the previous integer. Thereafter, each integer is 2 less than the previous integer. If the first integer and the last integer in the sequence are each 50 and the number of integers in the sequence is 29, m=?
A. 7
B. 9
C. 10
D. 12
E. 15
Source: Experts Global
A. 7
B. 9
C. 10
D. 12
E. 15
Source: Experts Global
The first m terms of the sequence form an increasing AP with A as 50 & d = 5
So mth term is given by A + (m - 1)*d ( formula for nth term of AP)
this is 50 +(m-1)5.
There is a second AP with the mth term of prev being it's first term hence A & d = -2.
We are given the last term of total sequence is also 50.... so we need to find the number of terms in the second AP.
mth term is 1st term, and we have 29 - m terms (since there are 29 terms in all)after that... hence 30 - m terms in all in second AP.
Hence the last term is the (30 - m)th term of 2nd AP. Applying the formula we get....
50 = A' + (n' - 1)d' .... A' is mth term from above 1st AP, d' = -2 and n' is (30 - m)
Plugging in the values:
\(50 = 50 +(m-1)5 + (30 - m - 1)*-2\)
\((29 - m)*2 = 5m - 5\) ....( 50's cancel and -2 term goes to the LHS)
\(58 + 5 = 7m\)
\(63 = 7m\)
\(m = 9\)
Hence Option (B) is correct.
Best,
Gladi
