Question

If the nth term of an A.P. is (3n + 1), then the sum of its first 'n' terms is
a)(2n^2 +4n)/3
b)(3n^2 +4n)/3
c)(3n² + 5n)/2
d)(3n^2 +4n)/2
​

Answer 1

Answer:

$\huge\frak{\fbox{Solution}}$

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).