Problem: Let Sn and Tn be the respective sums of the first n terms of two arithmetic series. If Sn:Tn=(7n+1):(4n+27) for all n , the ratio of the eleventh term of the first series to the eleventh term of the second series, is:
Answer Choices:
A. 4:3
B. 3:2
C. 7:4
D. 78:71
E. undetermined
Solution:
Let a1 and d1 be the first term and the common difference, respectively, of the first series, and let a2 and d2 be the first term and the common difference, respectively, of the second series.
Then TnSn=2a2+(n−1)d22a1+(n−1)d1=4n+277n+1.
Let u11 and v11, respectively, be the eleventh terms of the two series whose sums are Sn and Tn . Then
v11u11=a2+10d2a1+10d1=2a2+20d22a1+20d1. For n=21,TnSn=2a2+20d22a1+20d1.
Therefore, v11u11=4(21)+277(21)+1=111148=34