A sum of ` 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.
Step-by-Step Solution
Key Concept: For simple interest, the interest earned each year is constant because it is calculated only on the principal. A constant sequence is an arithmetic progression (AP) with common difference $d = 0$. The $n^{\text{th}}$ term of an AP is given by $a_n = a + (n-1)d$.
1. Interest for the first year
\[ I_1 = P \times \frac{R}{100} = 1000 \times \frac{8}{100} = 80 \text{ rupees} \]
2. Interest for subsequent years
Since the interest is *simple*, the same principal $P = 1000$ rupees is used every year. Hence
\[ I_2 = I_3 = \dots = I_n = 80 \text{ rupees} \]
The sequence of yearly interests is \[ 80,\,80,\,80,\,\dots \]
3. Check if the sequence is an AP
An AP has the form $a, a+d, a+2d, \dots$ where $a$ is the first term and $d$ the common difference. Here $a = 80$ and each successive term differs by $0$, i.e., $d = 0$. Therefore the interests do form an AP (a constant AP).
4. Use the AP formula to find the interest at the end of the 30th year
The $n^{\text{th}}$ term of an AP is \[ a_n = a + (n-1)d \]
Substituting $a = 80$, $d = 0$, and $n = 30$ gives
\[ a_{30} = 80 + (30-1)\times 0 = 80 \text{ rupees} \]
Hence the interest earned in the 30th year is Rs. 80.
5. (Optional) Total interest after 30 years
Total interest $= 30 \times 80 = 2400$ rupees.
Correct Answer: The interest each year is Rs. 80. These interests form an AP with first term $a=80$ and common difference $d=0$. Using $a_{30}=a+(30-1)d$, the interest at the end of the 30th year is Rs. 80 (total interest after 30 years = Rs. 2400).