Question

The ratio of the 1st 'n' terms of 2 APs is "(7n+1):(4n+27) ",then find the ratio of their 9th terms.
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Answer 1

Sum of the first n terms is given by

S = n/2 [2a + (n - 1)d]

Let the two sums be denoted by S & S*.

The ratio of these two sums can be written as

S/S* = [n/2 {2a + (n - 1)d}] / [n/2 {2a* + (n - 1)d*}]

= [2a + (n - 1)d] / [2a* + (n - 1)d*]

Now comparing it with the given terms , i.e ,

[2a + (n - 1)d] / [2a* + (n - 1)d*] = (7n + 1)/(4n + 27)

we see that coefficient of n is 7 in numerator and that of denominator is 4.

So d = 7 & d* =4.

Now we have two equations

2a + 7n - 7 = 7n + 1 & 2a* + 4n - 4 = 4n +27

Or , 2a = 8 & 2a* = 31

Or , a = 4 & a* = 31/2

Now the ratio of nth term of them , we get

T/T* = [a + (n - 1)d] / [a* + (n - 1)d*]

On putting the values of a , a* , d , d* & n = 9 , we get

T/ T* = 24/19...


Hope this will help u..!!!!