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Question
In the following situation, the sequence of numbers formed will form an A.P.?
Divya deposited ₹ 1000 at compound interest at the rate of 10% per annum. The amount at the end of first year, second year, third year, ..., and so on.
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Solution
Given: Divya deposited P = ₹ 1000 at compound interest r = 10% per annum.
Step-wise calculation:
1. General formula for the amount at the end of n years:
`A_n = P(1 + r/100)^n`
2. Here An = 1000(1.10)n. So
A1 = 1000 × 1.10
= ₹ 1100
A2 = 1000 × (1.10)2
= ₹ 1210
A3 = 1000 × (1.10)3
= ₹ 1331
3. Check successive differences:
A2 – A1 = 1210 – 1100
= ₹ 110
A3 – A2 = 1331 – 1210
= ₹ 121
Since 110 ≠ 121, the difference between consecutive terms is not constant.
The sequence of amounts 1100, 1210, 1331, ... is a geometric progression (each term multiplied by 1.10) and does not form an arithmetic progression (A.P.).
