The 4th and 10th terms of A.P. are 13 and 25 respectively. Find its (i) first term (ii) common difference (iii) 17th term.

We know that,

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

Given, 4th term is 13.

∴ a₄ = a + (4 - 1)d

⇒ 13 = a + 3d

⇒ a + 3d = 13 …….(1)

Given, 10th term is 25

∴ a₁₀ = a + (10 - 1)d

⇒ 25 = a + 9d

⇒ a + 9d = 25 ……(2)

Subtracting equation (1) from (2) we get,

⇒ a + 9d - (a + 3d) = 25 - 13

⇒ a + 9d - a - 3d = 12

⇒ 6d = 12

⇒ d = $\dfrac{12}{6}$

⇒ d = 2.

Substituting value of d in (1) we get,

⇒ a + 3(2) = 13

⇒ a + 6 = 13

⇒ a = 13 - 6

⇒ a = 7.

∴ a₁₇ = 7 + (17 - 1)2

= 7 + (16)2

= 7 + 32

= 39.

Hence, a = 7, d = 2 and a₁₇ = 39.