Question

find the AP whose third term is 16 and the seventh term exceeds its fifth term by 12
question no. 12

Answer 1

Given, a3 = 16

We know that a + (n-1) d 

Apply values in above formula, we get

a + (3 − 1) d = 16

a + 2d = 16  ------------------ (1)

Given, a7 − a5 = 12

a+ (7 − 1) d − a + (5 − 1) d = 12

(a + 6d) − (a + 4d) = 12

2d = 12

d = 6     --------------------------- (2)

Substitute equation (2) in (1), we get

= a + 2 * 6 = 16

= a + 12 = 16

a = 4

We know that ap = a,a+d,(a+2d),(a+3d)....

Substitute a,d values in above equation, we get

4,10,16,22..

Hope this helps!

Answer 2

a + 2d = 16 ------(1)

a + 4d + 12 = a + 6d ------(2)

from equ (2)

2d = 12

d = 6

on putting the value of d in equ (1) we get

a = 4

so the ap is 4 , 10 , 16 ,22 ...................