Given: The sum of n terms of two arithmetic progressions are in the ratio (7n + 1) : (4n + 27) . Let a1,d1 be the first term and common difference of the sequence whose sum of n term is given by Sn . Similarly, a2,d2 be the first term and common difference of the sequence whose sum of n term is given by Sn'. As we know that, Sn=
n
2
×[2a+(n−1)d] ⇒
Sn
Sn
=
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) Ratio of 11 th terms =
a1+10d1
a2+10d2
=
2a1+(21−1)d1
2a2+(21−1)d2
......(2) By substituting n = 21 in (1) we get ⇒
2a1+(21−1)d1
2a2+(21−1)d2
=
7(21)+1
4(21)+27
.......(3) From (2) and (3), we get Ratio of 11th terms = 148/111 Hence, the required ratio is 148 : 111