The fourth and seventh terms of an Arithmetic Progression (A.P.) are 60 and 114 respectively. Find the :

(a) first term and common difference.

(b) sum of its first 10 terms.

(a) Given,

The fourth and seventh terms of an Arithmetic Progression (A.P.) are 60 and 114 respectively.

We know that,

a_n = a + (n - 1)d

Fourth term :

⇒ a₄ = a + (4 - 1)d

⇒ a + 3d = 60 ….(1)

Seventh term :

⇒ a₇ = a + (7 - 1)d

⇒ a + 6d = 114 ….(2)

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

⇒ (a + 6d) - (a + 3d) = 114 - 60

⇒ a + 6d - a - 3d = 54

⇒ 3d = 54

⇒ d = $\dfrac{54}{3}$ = 18.

Substituting value of d = 18 in equation (1), we get :

⇒ a + 3d = 60

⇒ a + 3(18) = 60

⇒ a + 54 = 60

⇒ a = 60 - 54

⇒ a = 6.

Hence, first term = 6 and common difference = 18.

(b) Sum of the first 10 terms :

By formula,

$S_n = \dfrac{n}{2}[2a + (n - 1)d]$

Substituting values, we get :

$\Rightarrow S_{10} = \dfrac{10}{2}[2(6) + (10 - 1)18] \\[1em] = 5[12 + 162] \\[1em] = 5[174] \\[1em] = 870.$

Hence, sum of its first 10 terms = 870.