Question

the pth term of AP is (3p-1)/6.then sum of first n terms of the AP is. . ?

Answer 1

Answer:

Step-by-step explanation:

Sn=(n/2) [a(1)+a(n)]

= a(n)=(3-1)/6

= S(n)=(n/2) [(1/2)+(3n-1)/6]

= (n/2) [(3+3n-1)/6]

= (n/2) (3n+2)/6

= 3n^2+2n/12

= n/12(3n+1)

Answer 2

Answer:

$\frac{3n^{2}+n }{12}$

Step-by-step explanation:

Given:- $p^{th}$ term pf an A.P. = (3p - 1)/6.

To Find:- Sum of 1st n terms of A.P.

Solution:-

It is given that $p^{th}$ term = (3p - 1)/6.

1st term = (3 × 1 - 1)/6

             = (3-1)/6  

             = 2/6  = 1/3

2nd term = (3 × 2 - 1)/6

               = (6 - 1)/6  = 5/6

3rd term = (3 × 3 - 1)/6

              = (9 - 1)/6  = 8/6  = 4/3

nth term = (3 × n - 1)/6

              = (3n - 1)/6

As we know, Sum of all term = n/2 [1st term + last term]

⇒ $S_{n}$  =  $\frac{n}{2} [\frac{1}{3} + \frac{(3n - 1)}{6} ]$

⇒ $S_{n}$  =  $\frac{n}{2} [\frac{2 + 3n - 1}{6} ]$

⇒ $S_{n}$  =  $\frac{n}{2} [\frac{3n+1}{6} ]$

⇒ $S_{n}$  =  $\frac{3n^{2}+n }{12}$

Therefore, sum of 1st n terms of the A.P. is (3n² + n)/12.

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