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In a sequence of integers, each term in the first m integers is 5

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In a sequence of integers, each term in the first m integers is 5 [#permalink]

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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
[Reveal] Spoiler: OA

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Re: In a sequence of integers, each term in the first m integers is 5 [#permalink]

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pushpitkc wrote:
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


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.

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In a sequence of integers, each term in the first m integers is 5 [#permalink]

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New post 28 Mar 2018, 17:54
pushpitkc wrote:
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


term m needs to be even to subtract 2x and get 50
thus, m needs to be odd
looking at 7: term m=80; 80-50=15*2; 15+7≠29; no
looking at 9: term m=90; 90-50=20*2; 20+9=29; yes
B
In a sequence of integers, each term in the first m integers is 5   [#permalink] 28 Mar 2018, 17:54
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In a sequence of integers, each term in the first m integers is 5

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