If S_n = n(4n + 1) is the sum of n terms of an A.P., then its common difference is __________.

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Solution

Given:
$S_{n} = n (4 n + 1) Now, Sum of one term = S_{1} = 1 (4 + 1) = 5 \Rightarrow a_{1} = a = 5 . . . (1) Sum of two term s = S_{2} = 2 (8 + 1) = 18 \Rightarrow a_{1} + a_{2} = 18 . . . (2) Subtracting (1) from (2), we get a_{2} = 13 \Rightarrow a + d = 13 \Rightarrow d = 8 (∵ a = 5)$

Hence, if S_n = n(4n + 1) is the sum of n terms of an A.P., then its common difference is 8.

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If Sn = n(4n + 1) is the sum of n terms of an A.P., then its common difference is __________.

Given: Sn=n4n+1Now,Sum of one term=S1=14+1=5⇒a1=a=5 ...1Sum of two terms=S2=28+1=18⇒a1+a2=18 ...2Subtracting 1 from 2, we geta2=13⇒a+d=13⇒d=8 ∵ a=5 Hence, if Sn = n(4n + 1) is the sum of n terms of an A.P., then its common difference is 8.

If S_n = n(4n + 1) is the sum of n terms of an A.P., then its common difference is __________.