Question

The 7th term of an A.P is '32' and its 13th term is 62. Find the A.P.​

Answer 1

An = a+ (n-1)d

Now, a7= a+6d = 32---------------(1)

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

Subtracting (1) and (2)

-6d = (-30)

d = 5

Putting the value of 'd ' in (1)

a + 6 (5) = 32

a + 30 = 32

a = 32- 30

a = 2, d = 5

So, the required A.P is 2, 7, 12, 17,22.......

Answer 2

Step-by-step explanation :

7th term of A.P. (t_7) = 32

13th term of A.P. (t_13) = 62

To calculate :-

• A.P. = ?

Let's begin :-

As we know that general or nth term of an A.P. is calculated by the formula :

• tn = a + (n - 1) d

Here,

• a is first term
• d is common difference
• n is number of terms

t_7 = a + (7 - 1) d

t_7 = a + 6d

32 = a + 6d

a = 32 - 6d

And,

t_13 = a + (13 - 1) d

t_13 = a + 12d

62 = a + 12d

Substitute the value of a which we got above here.

62 = (32 - 6d) + 12d

62 = 32 - 6d + 12d

62 = 32 + 6d

6d = 62 - 32

6d = 30

d = 30 / 6

d = 5

Therefore, common difference is 5.

Now,

a = 32 - 6d

a = 32 - 6(5)

a = 32 - 30

a = 2

Therefore,

• A.P. is 2 , 7 , 12 ..... nth