Last visit was: 17 May 2024, 20:42 It is currently 17 May 2024, 20:42
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

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

SORT BY:
Tags:
Show Tags
Hide Tags
CEO
Joined: 26 Feb 2016
Posts: 2873
Own Kudos [?]: 5227 [19]
Given Kudos: 47
Location: India
GPA: 3.12
Director
Joined: 16 Sep 2016
Status:It always seems impossible until it's done.
Posts: 643
Own Kudos [?]: 2071 [2]
Given Kudos: 174
GMAT 1: 740 Q50 V40
GMAT 2: 770 Q51 V42
VP
Joined: 07 Dec 2014
Posts: 1071
Own Kudos [?]: 1576 [2]
Given Kudos: 27
Intern
Joined: 25 Sep 2017
Posts: 3
Own Kudos [?]: 0 [0]
Given Kudos: 303
Re: In a sequence of integers, each term in the first m integers is 5 [#permalink]
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

Bunuel

Hi Bunuel, can you please explain me this problem?

Thanks,
Nikita
Director
Joined: 16 Sep 2016
Status:It always seems impossible until it's done.
Posts: 643
Own Kudos [?]: 2071 [0]
Given Kudos: 174
GMAT 1: 740 Q50 V40
GMAT 2: 770 Q51 V42
Re: In a sequence of integers, each term in the first m integers is 5 [#permalink]
NikitaS wrote:

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,

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.

Best,
Math Expert
Joined: 02 Sep 2009
Posts: 93334
Own Kudos [?]: 624552 [4]
Given Kudos: 81898
Re: In a sequence of integers, each term in the first m integers is 5 [#permalink]
3
Kudos
1
Bookmarks
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

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 $$50 = [50 + 5(m - 1) - 2] + (-2)[(29 - m) - 1]$$:

$$0 = 5m - 7 -56 + 2m$$

$$7m=63$$;

$$m=9$$.

12. Sequences

For other subjects check:
ALL YOU NEED FOR QUANT.

Hope it helps.
Intern
Joined: 25 Sep 2017
Posts: 3
Own Kudos [?]: 0 [0]
Given Kudos: 303
Re: In a sequence of integers, each term in the first m integers is 5 [#permalink]

Intern
Joined: 25 Sep 2017
Posts: 3
Own Kudos [?]: 0 [0]
Given Kudos: 303
Re: In a sequence of integers, each term in the first m integers is 5 [#permalink]
Thanks Bunuel
Non-Human User
Joined: 09 Sep 2013
Posts: 33058
Own Kudos [?]: 828 [0]
Given Kudos: 0
Re: In a sequence of integers, each term in the first m integers is 5 [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Re: In a sequence of integers, each term in the first m integers is 5 [#permalink]
Moderators:
Math Expert
93334 posts
Senior Moderator - Masters Forum
3137 posts