An amount must be invested now to allow withdrawals of $1,200 per year for the n

Question

An amount must be invested now to allow withdrawals of $1,200 per year for the next 13 years and to permit $320 to be withdrawn starting at the end of year 5 and continuing over the remainder of the 13-year period as the $320 increases by 5% per year thereafter. That is, the withdrawal at EOY six will be $336.00, 5352.80 at EOY seven, and so forth for the remaining years. The interest rate is 12% per year Click the icon to view the interest and annuity table for discrete compounding when 5% per year. Click the icon to view the interest and annuity table for discrete compounding when i= 12% per year. The P amount is S-Round to the nearest dollar)

Explanation / Answer

This can be written as

-P+1200/(1+r)+1200/(1+r)^2+1200/(1+r)^3+1200/(1+r)^4+1200/(1+r)^5+1200/(1+r)^6+1200/(1+r)^7+1200/(1+r)^8+1200/(1+r)^9+1200/(1+r)^10+1200/(1+r)^11+1200/(1+r)^12+1200/(1+r)^13+320/(1+r)^5+320*1.05/(1+r)^6+320*1.05^2/(1+r)^7...

This should equal zero

So, 0=-P+-1071.428571 +956.6326531 + 854.1362974 + 762.6216941 + 680.9122269 + 607.9573454 + 542.8190584 + 484.6598736 + 432.73203 + 386.3678839 + 344.9713249 + 308.0101115 + 275.0090282+181.5765938 + 170.2280567 + 159.5888032 + 149.614503 + 140.2635965 + 131.4971218 + 123.2785516 + 115.5736422 +108.3502895

Hence, P=8988.23

Alternatively, -P+1200/1.12*(1-1/1.12^13)/(1-1/1.12)+320/1.12^5*(1-(1.05/1.12)^9)/(1-1.05/1.12)=0

Hence, P=8988.23

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