Question
Jo has started a savings plan with an investment of k1200 and is adding to this amount each month with an additional k120. If the account is paying 4.8% p.a interest on the increasing balance, find:
The time it will take to accumulate k10000 in this account
The time it will take to accumulate k10000 in this account
Answers
Answered by
GPT 3.5
Let n be the number of months it will take to accumulate k10000 in the account.
The increasing balance each month can be represented by the formula:
1200 + 120n
The account balance at the end of n months, including interest, can be calculated using the formula for compound interest:
10000 = (1200 + 120n)(1 + 0.048/12)^n
Simplifying the formula, we get:
10000 = (1200 + 120n)(1.004)^n
Expanding the equation further, we get:
10000 = 1200(1.004)^n + 120n(1.004)^n
Since we're looking for n, we can rearrange this equation to get it in the form of a polynomial:
8800 = 1200(1.004)^n + 120n(1.004)^n
Divide through by 120 to simplify:
73.3333 = (1.004)^n + n(1.004)^n
At this point, you can solve for n using trial and error or by using numerical methods (like Newton's method) to find the value of n that satisfies the equation.
The increasing balance each month can be represented by the formula:
1200 + 120n
The account balance at the end of n months, including interest, can be calculated using the formula for compound interest:
10000 = (1200 + 120n)(1 + 0.048/12)^n
Simplifying the formula, we get:
10000 = (1200 + 120n)(1.004)^n
Expanding the equation further, we get:
10000 = 1200(1.004)^n + 120n(1.004)^n
Since we're looking for n, we can rearrange this equation to get it in the form of a polynomial:
8800 = 1200(1.004)^n + 120n(1.004)^n
Divide through by 120 to simplify:
73.3333 = (1.004)^n + n(1.004)^n
At this point, you can solve for n using trial and error or by using numerical methods (like Newton's method) to find the value of n that satisfies the equation.