If the nth term of an A.P.is (3n + 1), then the sum of its first 'n' terms is
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Step-by-step explanation:
Tn=(3n-1)
Sn= sigma Tn = sigma (3n-1)
Sn= 3.sigma n - sigma(1)
Sn= 3.{n.(n+1)/2} - n
Sn= {3.n.(n+1)- 2n}/2. = n/2.(3n+1).
Putting n=p
Sp=p/2.(3p+1). . Answer.
Second-method:-
Tn= (3n-1) putting n=1,2,3,4…….
T1=3.1–1=2
T2=3.2–1=5
T3=3.3–1=8
T4=3.4–1=11. Thus the series is 2,5,8,11……. which is an A.P. in which a=2 , d=3 ,
n=p. , So=?
Sn=n/2.[2a+(n-1).d]
Sp= p/2[2.2+(p-1).3]
Sp=p/2.[4+3p-3]
Sp= p/2.(3p+1). Answer.
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