Find the sum of 12th term of the following AP series 7,11 and 15.

It is adding 4 each time.

Just calculate the first 12 terms and add them up.

Term 1 = 7 + 4*0 = 7

Term 2 = 7 + 4*1 = 11

Term 3 = 7 + 4*2 = 15

Term 4 = 7 + 4*3 = 19

Term 5 = 7 + 4*4 = 23

Term 6 = 7 + 4*5 = 27

Term 7 = 7 + 4*6 = 31

Term 8 = 7 + 4*7 = 35

Term 9 = 7 + 4*8 = 39

Term 10 = 7 + 4*9 = 43

Term 11 = 7 + 4*10 = 47

Term 12 = 7 + 4*11 = 51

Sum of first 12 terms = 348

Answer:the sum of the 12th term of the AP is 348

Step-by-step explanation:

7,11,15

Common difference=d=11-7

d=4

First term=7

Using the formula

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

S12=12/2 x (2x7+4x (12-1))

S12=6 x (14+4x11)

S12=6 x (14+44)

S12=6 x 58

S12=348