Question

If the ratio of the sum of the first n terms of two APs is (7n +1 ): (4n +27) , then find the ratio of 9th terms

Answer 1

Sn/S"n =(7n+1)/(4n +27)

=n/2{2.4 + ( n -1)7}/n/2{2.31/2+(n-1)4}

you can see that this is an standard form of Sn
e.g Sn = n/2{2a + ( n -1)d}

so, for Sn
a = 4 and d = 7

for S"n
a =31/2 and d = 4

now,

Tn/T"n ={a + ( n -1)d}/{a + ( n -1)d}

={4 + ( n -1)7}/{31/2 + ( n -1)4}

now,

T9/T"9 ={ 4 + 56}/{31/2 + 32}
={60 }/{95/2}

=120/95

=24/19

Answer 2

let a and d be the first term and the common difference of the 1st AP and let A and D be the first term and the common difference of the 2nd AP.so Sn/Sn, =7n+1/4n+26.remember that the n is same of the both APs .then,
2a+(n-1)d/2A+(n-1)D=7n+1/4n+27
or,putting 2n-1 in place of n in order to get the nth term.so we get,2a+(2n-2)d and 2A+(2n-2)D and we get a+(n-1)d/A+(n-1)D=7(2n-1)+1/4(2n-1)+27and for the 9 the term we get a+8d/A+8D=120/95=24/19