Question

A sum of Rs 1000 is invested at 8% per annum simple interest. Calculate the
Interest at the end of 1,2,3... years. Is the sequence of interest an AP? Find the
interest at the end of 30 years.​

Answer 1

Answer:2400

Explanation

P=Rs 1000,R=8% and T=1,2 and 3

we know that

Interest=P×R×T/100

Interest of 1 year=1000×8×1/100

=80

Interest of 2 year=1000×2×1/100

=160

Interest of 3 year=1000×3×1/100

=240

Interest at the end of 1,2,3 years are 80,160 and 240 respectively

Now,we have to find interest at the end of 30 years

a30=a+(n-1)d

=80+(30-1)80

=80+29×80

=2400

Interest at the end of 30 years= Rs 2400

Answer 2

Given:

$\implies$ P = Rs. 1000

$\implies$ R = 8%

$\implies$ T = 1, 2 and 3

We know that:

$\boxed{\sf{Interest = \frac{P \times R \times T }{100}}}$

By using this formula, we get:

Interest of 1 year:

$\implies$ $1000 \times 8/100 \times 1$

$\implies$ $80$

Interest of 2 years:

$\implies$ $1000 \times 8/100 \times 2$

$\implies$ $160$

Interest of 3 years:

$\implies$ $1000 \times 8/100 \times 3$

$\implies$ $240$

Hence:

Interest at the end of 1, 2 and 3 years are 80, 160 and 240 respectively.

So:

Interest at the end of 30 years:

$\implies$ $\sf{a_{n}= a + (n - 1)d}$

$\implies$ $\sf{a_{30} = a + (n - 1)d}$

$\implies$ $\sf{a_{30} = 80 + (30 - 1)80}$

$\implies$ $\sf{a_{30} = 80 + 29 \times 80}$

$\implies$ $\sf{a_{30} = 80 + 2320}$

$\implies$ $\sf{a_{30} = 2400}$

Therefore:

Interest at the end of 30 years is Rs. 2400