Part (1)
Equal-payment-series compound-amount factor
$(F/A, \, i, \, n) = \dfrac{(1 + i)^n - 1}{i}$
$(F/A, \, 2.5\%, \, 10) = \dfrac{(1 + 0.025)^{10} - 1}{0.025}$
$(F/A, \, 2.5\%, \, 10) = 11.2034$
Part (2)
Amount of investment after 10 years
$F = A(F/A, \, i, \, n)$
$F = 250,000(F/A, \, 2.5\%, \, 10)$
$F = \text{P}2,800,845.44$
Part (3)
Number of years for the investment to grow to P10,000,000.00
$F = \dfrac{A [ \, (1 + i)^n - 1 \, ]}{i}$
$10,000,000 = \dfrac{250,000 [ \, (1 + 0.025)^t - 1 \, ]}{0.025}$
$1.025^t = 2$
$t \log 1.025 = \log 2$
$t = 28.07 \text{ years}$
