Question

plz answer this question​

Answer 1

Step-by-step explanation:

Let the two series' be Tn and Tn′ with first terms a and a′ and common differences d and d′

The ratio of the sums of the series' Sn and Sn′ is given as,

Sn′Sn=[n/2][2a′+[n−1]d′][n/2][2a+[n−1]d]=4n+277n+1

Or,

a′+[(n−1)/2]d′a+[(n−1)/2]d=4n+277n+1 ...(1)

We have to find, 

T11′T11=a′+10d′a+10d

Choosing (n−1)/2=10 or n