Question

A sum of $5000$ is invested at $8\%$ simple interest per annum. Calculate the interest at the end of each year. Do these interests form an AP? Find the interest at the end of $30\text{yr}$.

Answer 1

Step 1: Find the interest at the end of each year:

$\text{Total Amount}=5000$

$\text{Simple Interest}=8\%$

Formula used: $\text{SI}=\frac{\text{PRT}}{100}$

$\begin{array}{rcl}\text{Interest at the end of 1st year}& =& \frac{5000\times 8\times 1}{100}\\ & =& 400\end{array}$

$\begin{array}{rcl}\text{Interest at the end of 2nd year}& =& a_2\\ & =& \frac{5000\times 8\times 2}{100}\\ & =& 800\\ \text{Interest at the end of 3rd year}& =& a_3\\ & =& \frac{5000\times 8\times 3}{100}\\ & =& 1200\end{array}$

Step 2: Check whether the sequence forms AP or not:

Here, the sequence is $400,800,1200,\dots$

Here, $a=400,a_1=800,a_2=1200$

$\begin{array}{cc}\Rightarrow & a_2-a_1=a_1-a\\ \Rightarrow & 400=400\end{array}$

Since the common difference is equal.

Therefore, the sequence forms AP.

Step 3. Find the interest at the end of $30\text{yr}$:

We know that the formula to find the $n\text{th}$ term of an AP is $a_n=a+\left(n-1\right)d$.

$\begin{array}{rcl}& \Rightarrow & a_{30}=400+\left(30-1\right)400\\ & =& 400+11600\\ & =& 12000\end{array}$

Therefore, Yes, the interest forms an AP and the interest at the end of $30\text{yr}$ is $12000.$