A sum of 2000 is invested at 6% simple interest per annum.

(i) Calculate the interests at the end of 1, 2, 3.... years.
(ii) Does the sequence of interests forms an A.P.?

(i) after t years, i = Prt = 2000(0.06)t = 120t

(ii) of course

Good but not sure if you 0

(i) To calculate the interest at the end of each year, we can use the formula for simple interest:

Interest = Principal × Rate × Time

For this problem, the principal (P) is $2000 and the rate (R) is 6% or 0.06. Now we just need to calculate the interest at the end of 1, 2, 3, etc. years.

Year 1:
Interest = $2000 × 0.06 × 1 = $120

Year 2:
Interest = $2000 × 0.06 × 2 = $240

Year 3:
Interest = $2000 × 0.06 × 3 = $360

Year 4:
Interest = $2000 × 0.06 × 4 = $480

And so on...

(ii) To determine if the sequence of interests forms an arithmetic progression (A.P.), we need to check if the differences between consecutive terms are constant.

Let's calculate the differences between the terms:

Difference between Year 2 and Year 1: $240 - $120 = $120
Difference between Year 3 and Year 2: $360 - $240 = $120
Difference between Year 4 and Year 3: $480 - $360 = $120

As we can see, the differences between consecutive terms are all $120, which is a constant value. Therefore, the sequence of interests forms an arithmetic progression (A.P.).

To calculate the interests at the end of each year, you need to use the formula for simple interest:

Interest = Principal × Rate × Time

In this case, the principal (P) is 2000 and the rate (R) is 6% or 0.06.

(i) Now, let's calculate the interests at the end of 1, 2, 3.... years:

For year 1: Interest = 2000 × 0.06 × 1 = 120

For year 2: Interest = 2000 × 0.06 × 2 = 240

For year 3: Interest = 2000 × 0.06 × 3 = 360

(ii) To determine whether the sequence of interests forms an arithmetic progression (A.P.), we need to check if the differences between consecutive terms are constant.

Let's calculate the differences:

Difference between the 2nd and 1st term: 240 - 120 = 120
Difference between the 3rd and 2nd term: 360 - 240 = 120

Since the difference between consecutive terms is constant (120), the sequence of interests forms an arithmetic progression (A.P.).