Question

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

1. first term and common difference.
2. sum of its first 10 terms.

Answer 1

(a) 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 = `54/3​`

= 18

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

⇒ a + 3d = 60

⇒ a + 3(18) = 60

⇒ a + 54 = 60

⇒ a = 60 − 54

⇒ a = 6

(b) Sum of the first 10 terms:

By formula,

`S_n = n/2[2a + (n − 1)d]`

`S_n​ = 2n​[2a + (n − 1)d]`

Substituting values, we get:

⇒ `S_10 = 10/2[2(6)+(10−1)18]`

= 5[12 + 9 × 18]

= 5[12 + 162]

= 5[174]

= 870