Question

Find the sum of 16 terms of an AP where nth term is 3n – 1.

step by step plzz asap...

Answer 1

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.

Answer 2

Answer:

392

Step-by-step explanation:

First term,a= 3(1)-1=3-1=2

So a=2

and similarly 2nd term is 3(2)-1=6-1=5

We know 2nd term is a+d=5=2+d

d=5-2=3

We also know that Sum of n terms of an A.P = n/2 *(2a+(n-1)d)

So sum of 16 terms is 16/2 * (2*2+15*3)

8*(4+45)

8*49

392