Question

To accumulate 8000 at the end of 3n years, deposits of 98 are made at the end of each of the first n years and 196 at the end of each of the next 2n years. The annual effective rate of interest is i. You are given (1 ) 2.0 n + = i .

Answer 1

Answer: $i = 12.25%$

Step-by-step explanation:

The accumulative value for n years will be:

$98*((n+i)^{n}-1)/i$

$=98*((2-1)/i$

$=98*(1)/1$

=$\frac{98}{i}$

The accumulative value for 2n years will be:

$((1+i)^{2n}-1)/i$

=4* 98/i

=$\frac{392}{i}$              .....called it equation 1

The accumulative value for 3n years will be:

$196*((n+i)^{n}-1)/i$

$=196*((4-1)/i$

=$\frac{196*3}{i}$

=$\frac{588}{i}$                   ....called it equation 2

Now sum equation 1 and equation 2 together

$\frac{392}{i}$+$\frac{588}{i}$=8000

$980i=8000$

divide both side by 980 to get i

i = $\frac{8000}{980}$

$i = 12.25%$