Question

The 7th term of AP 32 and it's 13th term is 62 than find ap

Answer 1

T7 = 32

=> a + 6d = 32 ------(1)

T13 = 62

=> a + 12d = 62 ------(2)

On subtracting equation 1 from 2, we get

6d = 30

=> d = 5

Now,

On substituting the value of d in equation 1, we get

a + 6 × 5 = 32

=> a + 30 = 32

=> a = 2

Required AP is :-

2, 7, 12,.......

Answer 2

a7 = 32
a + 6d = 32...........(1)

a13 = 62
a+ 12d = 62 ............(2)

from equation (1)and (2)
6d = 30
d = 5

put the value of d in equation (1)
a + 6× 5 = 32
a + 30 = 32
a = 2

so the AP is
2 , 7 , 12 , 17........