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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