Question

A manufacturer of TV sets produces 600 units in the third year and 700 units in the $7^{th}$ year. Assuming that the production increases uniformly by a fixed number every year, find :

(i) the production in the first year.

(ii) the production in the ${10}^{th}$ year.

(iii) the total production in 7 years.

Answer 1

Since production increases by a fixed number every year, it is an AP.

Given 3rd year production is 600.

So, $a_3$ = 600

We know, $a_n$ = a + (n - 1) d

$a_3$ = a + (3 - 1)d
600 = a + 2d
600 - 2d = a ................eq (1)

Given 7th year production is 700.

So, $a_7$ = 700

We know, $a_n$ = a + (n - 1) d

$a_7$ = a + (7 - 1)d
700 = a + 6d
700 - 6d = a ................eq (2)
From (1) & (2)
600 - 2d = 700 - 6d
-100 = -4d
d = 25
Putting the value of d in eq(1)
a = 600 - 2d
a = 600 - 2(25) = 600 - 50
a = 550
(i) So, production in 1st year is 550 sets.

(ii) We need to find the production in the 10th year.
i.e., $a_{10}$
$a_{10}$ = 550 + (10 - 1)(25) = 550 + (9) (25) = 550 + 225 = 775

(iii) We need to find the production in first 7 years.
i.e., $S_7$
a = 550, d = 25 and n = 7
$S_7$ = $\frac{n}{2}$ [2a + (n - 1) d]

$S_7$ = $\frac{7}{2}$ [2 x 550 + (7 - 1) (25)]

$S_7$ = $\frac{7}{2}$ [1100 + (6) (25)]

$S_7$ = $\frac{7}{2}$ [1100 + 150]
$S_7$ = $\frac{7}{2}$ (1250) = 4375