Engineering Board Exam Reviewer · Uniform Gradient Series · hard
A deposit of P10,000 is made at the end of year 1 and each later deposit is 10% larger than the previous one, for 10 years. Find the present worth at 10% interest.
- P100,000
- P61,445.67
- P110,000
- P90,909.09
Answer
D. P90,909.09
Solution
Here the growth rate equals the interest rate (g=i), so each term has present worth A1/1+i. Hence P=nA1/1+i= 10(10,000)1.10=P90,909.09. (The usual formula would divide by i-g=0.)
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