Question

If the sum of first n terms of two arithmetic progression are in the ratio (7n + 1) : (4n + 27), find the ratio of their 11th terms.

Answer 1

similar to this.
http://openstudy.com/study#/updates/50b0a39de4b0e906b4a5fe9d

Answer 2

very long answer..eh aslo similar http://www.answerbag.com/q_view/1128650

Answer 3

oh!!!

Answer 4

ok @coolshubhs

Answer 5

Let the two A.Ps be A.P1 and A.P2
and let the first term and common differenceof A.P1 be a1and d1.
and let the first term and common differenceof A.P2 be a2and d2.
now. According to question
Sum of n terms of A.P1/Sum of n terms of A.P2 = (7n+1)/(4n+27)
n/2 [2a1 + (n-1)d1] / n/2 [2a2 + (n-1)d2] = (7n+1)/(4n+27)
[2a1 + (n-1)d1] / 2a2 + (n-1)d2] = (7n+1)/(4n+27) ...........(1)


Now Ratio of their 11th term.
a1 + (11-1)d1/a2 + (11-1)d2
multiply and divide by 2
2a1 + 20d1/2a2 + 20d2..................(2)

Comparing (1) and (2)
n-1=20
n=21

therefore
Ratio of their 11th term.
7(21) + 1 / 4(21) + 27
148/111

Read more: If the sum of first n terms of two arithmetic progression are in the ratio (7n + 1) : (4n + 27), find the ratio of their 11th terms.

Answer 6

but Comparing (1) and (2) ???

Answer 7

??

Answer 8

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