If the ratio of the sum of the first n
terms of two A.Ps is (7n + 1) : (4n +
27), then find the ratio of their 9th
terms.
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Let first AP be a1,a2,a3...
Let first term be A and common difference be D.
Sum of n terms= n/2(2A+(n-1)D)
Second AP be b1, b2, b3,...
Let first term be b and common difference be d.
Sum of n terms= n/2(2b +(n-1)d)
sum of n terms of first AP = 2A+(n-1)D
sum of n terms of second AP 2b+(n-1)d
n/2 and n/2 cancel out above
2A+(n-1)D = 7n+ 1
2b+(n-1)d 4n+27
Replace n by 2n-1,
So that it becomes,
2A+(2n-1-1)D
=> 2(A+(n-1)D)
similarly,
=>2(b+(n-1)d)
(A+(n-1)D) = 7(2n-1)+ 1
(b+(n-1)d) 4(2n-1)+27
The required ratio can be found by putting n=9
on RHS.
7(17) + 1
4(17)+27
=> 120
95
=====> that is 24
15
Ratio of their 9th terms is 24/15.
rickmorgan:
pls mark as brainliest... spent a lotta time on this...
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