NCERT Class X Chapter 5: Arithmetic Progression Example 9
NCERT Class X Chapter 5: Arithmetic Progression Example 9
Question:
A sum of Rs 1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years making use of this fact.
Given:
Principal (P) = Rs 1000
Rate of interest (R) = 8% per year
To Find:
Interest at the end of each year.
Whether the interests form an AP.
Interest at the end of 30 years.
Formula:
Simple Interest (SI) =
P × R × T
100
where,
P is principal,
R is rate, and
T is time in years.
Solution:
Interest at the end of 1st year (T=1) = 1000 × 8 × 1 100 = Rs 80
Interest at the end of 2nd year (T=2) = 1000 × 8 × 2 100 = Rs 160
Interest at the end of 3rd year (T=3) = 1000 × 8 × 3 100 = Rs 240
The interest after at the end of each year is 80, 160, 240, ....
Let a1 = 80, a2 = 160, a3 = 240
d = a2 - a1
⇒ d = 160 - 80 = 80
Also, 240 - 160 = 80
Therefore the interests form an AP with first term a = 80 and common difference d = 80.
Interest at the end of 30 years = a + (n-1)d Here, n = 30
⇒ Interest at the end of 30 years = 80 + (30-1)80
⇒ Interest at the end of 30 years == 80 + 29 × 80
⇒ Interest at the end of 30 years == 80 + 2320
⇒ Interest at the end of 30 years == Rs 2400
Result:
The interest at the end of 30 years is Rs 2400.
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