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Exercises · Q13

Q.A person deposits ₹10,000 at the end of each year for 2 years into a fund earning 5% per annum compounded annually. Find the accumulated value. [Given (1.05)^2 = 1.1025]

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End-of-year deposits, so this is an immediate annuity with C=10000C=10000, i=0.05i=0.05, n=2n=2:

A=C[(1+i)n−1i]=10000[(1.05)2−10.05]=10000[1.1025−10.05]=10000×0.10250.05=10000×2.05=₹20,500.A=C\left[\frac{(1+i)^{n}-1}{i}\right]=10000\left[\frac{(1.05)^{2}-1}{0.05}\right]=10000\left[\frac{1.1025-1}{0.05}\right]=10000\times\frac{0.1025}{0.05}=10000\times2.05=₹20{,}500.

Check (independent verification) — grow each deposit separately: …

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