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how many in an ap

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how many in an ap

by maihuna » Sun May 10, 2009 8:07 am
Between 1 and 31, m numbers have been inserted in such a way that the resulting
sequence is an A. P. and the ratio of 7th and (m – 1)th numbers is 5 : 9. Find the
value of m.

11
12
13
14
15
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Re: how many in an ap

by dtweah » Sun May 10, 2009 12:43 pm
maihuna wrote:Between 1 and 31, m numbers have been inserted in such a way that the resulting
sequence is an A. P. and the ratio of 7th and (m – 1)th numbers is 5 : 9. Find the
value of m.

11
12
13
14
15
The seventh term of the sequence is question is
1+6d, where d is the arithmetic difference. The problem does not restrict to integers.
It may be that finitely many d can produce a 7th term which meets the parameters of the problem, ie that the sequence stays between 1 and 31.

(1+6d)/M=5/9, where M is the (m-1)th of the sequence.

M=9(1+6d)/5

So we know the m-1 term has to be a multiple of 5.

Let d= 2/3

Then the 7th term is 5
And the M-1 term is 9

Now 9 is what term of the sequence?
9=1+(n-1)2/3
27=3+2n-2
N=14.
So 14=m-1
M=15.

Choose E

I am not certain another d cannot produce a similar result. I leave that for further verification.
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by sureshbala » Sun May 10, 2009 7:47 pm
First of all I think there's a small correction in the question.

It has to be "the ratio of the 7th term and the (m-1)th term of the inserted numbers are in the ratio 5:9


If "n" numbers are inserted between A and B (A<B) such that the resulting numbers are in AP, it is evident that the common difference of the series will be (B-A)/(n+1).

(Since there are n+2 numbers and the total available difference is B-A, it must be divided into n+1 equal parts)

So coming to our problem, the common difference is 30/m+1.

So from the options first choose the number m such that m+1 divides 30. Here in the given options there is only one value i.e. m = 14.

If m=14, then common difference d = 2.

Now 7th term of the m numbers inserted = 8th term of the entire series = 1+7x2 = 15.


And the (m-1)th i.e. 13th term of the m numbers inserted = 14th term of the entire series = 1+13x2 = 27.

Hence their ratio is 15:27 = 5:9.

Thus M must be 14.
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by dtweah » Mon May 11, 2009 2:44 am
sureshbala wrote:First of all I think there's a small correction in the question.

It has to be "the ratio of the 7th term and the (m-1)th term of the inserted numbers are in the ratio 5:9


If "n" numbers are inserted between A and B (A<B) such that the resulting numbers are in AP, it is evident that the common difference of the series will be (B-A)/(n+1).

(Since there are n+2 numbers and the total available difference is B-A, it must be divided into n+1 equal parts)

So coming to our problem, the common difference is 30/m+1.

So from the options first choose the number m such that m+1 divides 30. Here in the given options there is only one value i.e. m = 14.

If m=14, then common difference d = 2.

Now 7th term of the m numbers inserted = 8th term of the entire series = 1+7x2 = 15.


And the (m-1)th i.e. 13th term of the m numbers inserted = 14th term of the entire series = 1+13x2 = 27.

Hence their ratio is 15:27 = 5:9.

Thus M must be 14.
If m=14, then common difference d = 2.

Thanks Sura for the rich analysis. Your correction transformed the problem and provided further insights into such problems. I always thought there was a troubling relationship between the m and the terms in the original posing.
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