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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

50+(m-1)*5=50+(29-m)*2
m=9
B
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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


Bunuel

Hi Bunuel, can you please explain me this problem?

Thanks,
Nikita
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Bunuel

Hi Bunuel, can you please explain me this problem?

Thanks,
Nikita

Hi NikitaS,

Can you please specify which part is troubling you? I have provided with a detailed solution as follows...

Kindly go through it once and let me know. I would be happy to help :-)

Best,
Gladi

Gladiator59
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

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
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Hi Gladiator59 ,

Thank you for your reply...I missed to see your explanation!! Thanks for your help. :-)
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Thanks Bunuel :-)
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