NCERT Class X Chapter 5: Arithmetic Progression Example 16(iii)
NCERT Class X Chapter 5: Arithmetic Progression Example 16(iii)
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 total production in first 7 years
Given:
Production in the 3rd year = 600 sets
Production in the 7th year = 700 sets
To Find:
Total production in the first 7 years
Formula:
Arithmetic progression: an = a1 + (n-1)d , Sn = n 2 (a1 + an)
Solution:
Let an be the production in the nth year.
We are given that a3 = 600 and a7 = 700.
Since the production increases uniformly, this forms an arithmetic progression.
Using the formula an = a1 + (n-1)d, we have:
a3 = a1 + 2d = 600
⇒ a1 + 2d = 600 (1)
a7 = a1 + 6d = 700
⇒ a1 + 6d = 700 (2)
Subtracting (1) from (2):
4d = 100
⇒ d = 25
Substituting d = 25 in (1):
a1 + 2(25) = 600
⇒ a1 = 550
Now we find the total production in the first 7 years using the formula:
⇒ Sn = n 2 (a1 + an)
⇒ S7 = 7 2 (550 + 700)
⇒ S7 = 7 2 (1250)
⇒ S7 = 7(625)
⇒ S7 = 4375
Result:
Total production in first 7 years = 4375
Next question solution:
NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 1(i)
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