Question

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

Answer 1

a= first term
d= common difference

nth term = a + (n-1)d
7th term = a + 6d = 32 ..............1

13th term = a+ 12d =62 ............. 2

subtracting 1 from 2, we get
d=5
substituting the value of d in equation 1,
we get, a = 2

hence
the A. P. is..
2, 7, 12, 17, 22, 27, 32, .....

Answer 2

Answer:

A. P. is 2, 7, 12, 17, 22, 27, 32, .....

Step-by-step explanation:

Given :  

a7 = 32,  a13 = 62

 

Let 'a' be the first term and common difference be 'd'.

nth term , an = a + (n -1)d

a7 = a + (7 - 1)d

32 = a + 6d …………..(1)

 

a13  = a + (13 - 1)d

62 = a + 12d…………..(2)

 

On subtracting equation (1) from (2)

(a + 12d) - (a + 6d)  = 62 - 32

a + 12d - a - 6d  = 30

a - a + 12d - 6d = 30

6d = 30

d = 30/6

d = 5

 

On substituting the value of d = 5  in eq 1,

32 = a + 6d

32 = a + 6 × 5  

32 = a + 30

a = 32 - 30

a = 2

First term , a = 2  

Second term , a2 = (a + d) =  2 + 5 = 7

Third term , a3 = (a + 2d) = 2 + 2×5 = 2 + 10 = 12  

Fourth term , a4 = (a + 3d) = 2 + 3×5

a4 = 2 + 15 = 17

Fifth term ,a5 = (a + 4d) = 2 + 4 × 5 = 2 + 20 = 22

Hence , A. P. is 2, 7, 12, 17, 22, 27, 32, .....

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