An account is credited interest using 6% simple interest rate from the date of each deposit into > the account. Annual payments of 100 are deposited into the account. Calculate the accumulated value of the account immediately after the 20th deposit.
I recognize that this is like an annuity-immediate, but whereas if I were to use the normal accumulated value formula for an annuity-immediate interest is normally compounded, this problem only uses simple interest.
I'm also pretty sure that I know how to solve this problem the long way
- 100(1.06) = 106
- 200(1.06) = 212 + 6 (this is the interest earned the first year) = 218
- 300 (1.06) = 318 + 18 (this is the 6 in interest from the first year and the 12 in interest from the 2nd year) = 336
I could continue this pattern for 20 terms but this would be time consuming. Is there a faster way to solve this problem?
With simple interest, the future value of a payment of $1$ that has accrued interest over $t$ years at a rate of $i$ per year is $1 + it$, whereas in compound interest, it is $(1+i)^t$.
In an annuity, regardless of how interest is accrued, the accumulated value can still be calculated by accumulating each individual payment up to the valuation point. So the idea here is to find the accumulated value of a representative payment, and then take an appropriate sum over all such payments.
The last payment has had no time to accrue interest, so its accumulated value is equal to $100$.
The second-to-last payment has had $1$ year to accrue interest, so its accumulated value is $100(1+0.06) = 106$.
The third-to-last payment has had $2$ years to accrue interest, so its accumulated value is $100(1+(0.06)(2)) = 112$.
In general, the $k^{\rm th}$-to-last payment has had $k-1$ years to accrue interest and its accumulated value is $100(1+(0.06)(k-1))$.
Since there are $20$ payments in total, the accumulated value of the full series of payments is $$\sum_{k=1}^{20} 100(1 + 0.06(k-1)) = 100 \left(20 + 0.06 \sum_{k=1}^{20} (k-1) \right) = 100 \left( 20 + 0.06 \frac{19(20)}{2}\right) = 3140.$$