Question

the 7th term of an A.P is 32 and 13th term is 62. Find the A.P.

Answer 1

Hi ,

Let a and d are the first term and

common difference of an A.P

a , a + d , a + 2d , ... is an A.P


we know that ,

n th term = an

an = a + ( n - 1 )d-----( 1 )

According to the problem given,

a7 = 32

a + ( 7 - 1 )d = 32

a + 6d = 32-----( 2 )

a13 = 62

a + ( 13 - 1 )d = 62

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

Subtract ( 1 ) from ( 2 ) , we get

6d = 30

d = 30/ 6

d = 5

Put d = 5 in equation ( 1 ) ,

a + 6d = 32

a + 6 × 5 = 32

a + 30 = 32

a = 32 - 30

a = 2

Therefore ,

a = 2 and d = 5 ,

Required A.P is

a , a + d , a + 2d , a + 3d , ...

2 , 2 +5 , 2 + 2 ×5 ,..

2 , 7 , 12 , 17 ,...is required A.P

I hope this helps you.

:)

Answer 2

Here is your solution

a+6d=32, .............. (1)
a+12d=62................... (2)

on subtracting equation 1 from. ii

a+12d-a-6d=62-32
6d= 30
d=5

hence, a=2

Ap are 2,2+5,2+5+5,2+5+5+5

Ap are 2,7,12,17

hope it helps you