Question

A manufacturer of T.V. sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that
the production increases uniformly by a fixed number every year, find
(i) The production in the 1 year (ii) The production in the 10 year​

Answer 1

123455678900sddgjiyrdxbjuyd

Answer 2

Step-by-step explanation:

Let, increased production every year is d. Since it is same, therefore the production is increasing in arithmetic progression. Let first year production is a.

$∴ \: \: \: \: \: \: \: \: \: \: a_{3} = 600 \\ \: \: \: \: \: \: \: \: a_{7} = 700$

$Therefore, from \: a_{n} = a + (n - 1)d,$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: a_{3} = a + (3 - 1)d = 600$

$⇒\: \: \: \: \: \: \: \: \: \: a + 2d = 600 \: \: \: \: \: \: - - (1)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: a_{7} = a + (7 - 1)d = 700$

$⇒ \: \: \: \: \: \: \: \: \: a + 6d = 700$

From eqns. (1) and (2)

$a + 2d = 600 \\ a + 6d = 700 \\ - \: \: \: \: \: \: \: - \: \: \: \: \: \: \: - \\ - 4 = - 100 \: \: \: \: \: \: \: \: \: \\ (on \: subtracting)$

$∴ \: \: \: \: \: \: \: \: \: \: \: d = \frac{ - 100}{ - 4} = 25$

From eqn. (1),

$\: \: \: \: \: \: \: a + 2 \times 25 = 600$

$⇒ \: \: \: \: \: \: \: \: \: a = 600 - 50 \\ \: \: \: \: \: \: \: \: \: = 550$

∴ production of first year = 550 T. V.

$</p><p>Production \: in \: {9}^{th} \: year \: a_{9} = a + 8d$

$\: \: \: \: \: \: \: \: \: \: \: \: = 550 + 8 \times 25$

$\: \: \: \: \: \: \: \: \: \: \: \: = 550 + 200$

$\: \: \: \: \: \: \: \: \: \: \: \: = 750 \: T.V.$

Production in first 7 years

$From, S_{n} = \frac{n}{2} [2a + (n - 1)d]$

$S _{7} = \frac{7}{2} [2 \times 550 + (7 - 1).25]$

$\: \: \: \: \: \: \: \: \: \: = \frac{7}{2} (1100 + 6 \times 25)$

$\: \: \: \: \: \: \: = \frac{7}{2} (1100 + 150)$

$\: \: \: \: \: \: \: \: = \frac{7}{2} \times 1250 \\ \: \: \: \: \: \: \: \: = 7 \times 625 \\ \: \: \: \: \: \: \: = 4375.$