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AP ratio

by maihuna » Sun May 10, 2009 6:54 am
The sum of n terms of two arithmetic progressions are in the ratio
(3n + 8) : (7n + 15). Find the ratio of their 12th terms.

7/16
9/16
11/23
16/9
16/7
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by scoobydooby » Sun May 10, 2009 11:07 am
ratio of the sum of the nth term of the two APs:

=n/2[2a1+(n-1)d1]/n/2[2a2+(n-1)d2]

=>[2a1+(n-1)d1]/[2a2+(n-1)d2]=(3n+8)/(7n+15)............1

we want a1+11d1/a2+11d2
so we manipulate the LHS of 1 to get the ratio in the above form by putting n-1=22 or n=23

2(a1+11d1)/2(2a2+11d2)=3*23+8/7*23+15
=(a1+11d1)/(a2+11d2)=69+8/151+15
=>ratio of the 12th terms=77/176=7/16

hence, A
Last edited by scoobydooby on Thu May 14, 2009 8:07 am, edited 1 time in total.
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by kanha81 » Thu May 14, 2009 7:07 am
scoobydooby wrote:ratio of the sum of the nth term of the two APs:

=n/2[2a1+(n-1)d1]/n/2[2a2+(n-1)d2]

=>[2a1+(n-1)d1]/[2a2+(n-1)d2]=(3n+8)/(7n+15)............1

we want a1+11d1/a2+11d2
so we manipulate the LHS of 1 to get the ratio in the above form by putting n-1=22 or n=23

2(a1+11d1)/2(2a2+11d2)=3*23+8/7*23+15
=(a1+11d1)/(a2=11d2)=69+8/151+15
=>ratio of the 12th terms=77/176=7/16

hence, A
I am quite confused!

How did you manipulate and came to know that n-1=22?
From what I understand, we need to find the ratio for the 12th term.

So then, shouldn't the equation form be-

(3n+5) / (7n+15) = [2a1+(n-1)d1] / [2a2+(n-1)d2]
where n=12, so rhs becomes- (2a1+11d1) / (2a2+11d2)
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by scoobydooby » Thu May 14, 2009 8:06 am
we have (3n+5) / (7n+15) = [2a1+(n-1)d1] / [2a2+(n-1)d2]
=>(3n+5) / (7n+15) =[a1+(n-1)d1/2]/[a2+(n-1)d2/2]

(divding numerator and the denominator of the RHS by 2)

for the ratio of the 12th terms we need: (a1+11d1) / (a2+11d2)

by observing the RHS, we can obtain this by putting (n-1)/2=11
=>n=22+1=23
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