Question

Find the ap whose 3rd term is 16 and 7th term exceeds the 5th term by 11

Answer 1

Answer:

Step-by-step explanation:

Hope it will help

Answer 2

3rd term=a+2d=16-[1]

7th term-5th term=11[2]

a+6d-a+4d=(+a and -a will be cancelled)

2d=11

d=5.5

substitute d=5.5 in[1]

a+2(5.5)=16

a=5

so AP= a=5

a+d=10.5

a+2d=16

hence it is proved