Question

DA
the 4th and 10th terms of an AP are 13 and 25
respectively. Find the first term and the common
sibb erence of the Ap. Also find its 17th term​

Answer 1

Answer:-

Given:

4th term of an AP = 13

10th term = 25

We know that,

nth term of an AP [ a(n) ] = a + (n - 1)d

Hence,

a(4) = 13

→ a + (4 - 1)d = 13

→ a + 3d = 13 -- equation (1)

a(10) = 25

→ a + (10 - 1)d = 25

→ a + 9d = 25 -- equation (2)

Subtract equation (1) from (2).

→ a + 9d - (a + 3d) = 25 - 13

→ a + 9d - a - 3d = 12

→ 6d = 12

→ d = 12/6

→ d = 2

Putting the value of d in equation (1) we get,

→ a + 3 * 2 = 13

→ a + 6 = 13

→ a = 13 - 6

→ a = 7

Now,

a(17) = a + (17 - 1)d

Putting the values of a and d we get,

→ a(17) = 7 + 16 * 2

→ a(17) = 39

Therefore,

• a (first term) = 7

• d (common difference) = 2

• 17th term = 39.

Answer 2

Step-by-step explanation:

The 4th and 10th terms of an AP are 13 and 25

respectively.

an = a + (n - 1)d

a4 = a + (4 - 1)d

13 = a + 3d

13 - 3d = a ...........(1)

a10 = a + (10 - 1)d

25 = a + 9d

25 - 9d = a .........(2)

On comparing (1) & (2) we get,

13 - 3d = 25 - 9d

6d = 12

d = 2

Substitute value of d in (1)

13 - 3(2) = a

13 - 6 = a

7 = a

Hence, the common difference is 2 and first term is 7.

Now,

a17 = a + (17 - 1)d

a17 = 7 + 16(2)

a17 = 7 + 32

a17 = 39

Hence, the 17th term is 39.