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)d2​2a1​+(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
v11​u11​​=a2​+10d2​a1​+10d1​​=2a2​+20d2​2a1​+20d1​​. For n=21,TnSn​=2a2​+20d2​2a1​+20d1​​.
Therefore, v11​u11​​=4(21)+277(21)+1​=111148​=34​