Question

Find the sum of first 11 terms of the AP:3,8,13,........?​

Answer 1

Answer:

S11=308

Step-by-step explanation:

Given:-

a=3,d=5,n=11.

To Find:-

S11

Formula Applied:-

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

Solution:-

S11=11/2[2(3)+(11-1)(5)]

S11=11/2[6+(10)(5)]

S11=11/2(6+50)

$s11 = \frac{11}{2} \times 56$

S11=11×28

S11=308

Answer 2

a1=3, a2=8

common difference (d) = a2-a1=8-3=5

an= a+(n-1)d

a11=3+(11-1)5

a11=3+50

a11=53

Now, the sum of first 11 terms of this AP

Sn=n/2(a+an)

S11=11/2(3+53)

$S11 = \frac{11}{2} \times 56 =30 8$

Remember:

an= l (last term)