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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
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