NCERT Class X Chapter 5: Arithmetic Progression Example 16(ii)

Question:

A manufacturer of TV 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 : the production in the 10th year

Given:

Production in the 3rd year = 600 sets
Production in the 7th year = 700 sets

To Find:

Production in the 10th year

Formula:

Let the production in the nth year be given by the arithmetic progression: a_n = a₁ + (n-1)d, where a₁ is the production in the first year and d is the common difference.

Solution:

Let a₃ = 600 and a₇ = 700. Then we have:

a₃ = a₁ + 2d = 600 ⇒ a₁ = 600 - 2d

a₇ = a₁ + 6d = 700

Substituting a₁ = 600 - 2d:

(600 - 2d) + 6d = 700

4d = 100

d = 25

Now, we find a₁:

a₁ = 600 - 2(25) = 600 - 50 = 550

Finally, we find a₁₀:

a₁₀ = a₁ + 9d = 550 + 9(25) = 550 + 225 = 775

Result:

The production in the 10th year is 775 sets.

Next question solution:

NCERT Class X Chapter 5: Arithmetic Progression Example 16(iii)

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