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An investment of 1000 accumulates to 1360.86 at the end of five years. If the force of interest is δ during the first year and 1.5δ in each subsequent year. How to find δ?

fae
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2 Answers2

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You need to solve the equation

$$1000 (1+\delta)(1+1.5\delta)^4=1360.80$$

(I get $4.54225\%$)

Patricio
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You missed the integral at the exponent.

$$1000\cdot e^{\int_0^1 \delta \ ds +4\cdot \int_0^1 1.5\cdot \delta \ ds}=1360.86$$

Dividing the equation by 1000 the equation becomes

$$e^{\int_0^1 \delta \ ds +4\cdot \int_0^1 1.5\cdot \delta \ ds}=1.36086$$

$$e^{7\cdot \delta}=1.36086$$ $$7\cdot \delta=\ln(1.36086)$$ $$\delta=\frac{1}{7}\cdot \ln(1.36086)\approx 0.04401669$$

It seems that you haven´t omitted the integrals at your calculation if I compare our results. So I can confirm your result. Can you now find the equivalent annual effective interest rate in the first year?

callculus42
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  • (1+i)^5 = e^(0.04401669326)(5) and the effective rate i got is 0.04499979914 – fae May 06 '21 at 16:04
  • Only for the fiirst year. But at the end your result is right. $$(1+i)=e^{\Large{\int_0^1 0.0440166932 \ d \delta } }=1.045$$ – callculus42 May 06 '21 at 16:20
  • what does the 1.045 value for – fae May 06 '21 at 16:51
  • @fae This is $1+i$. Therefore the annual interest rate is $i=0.045=4.5%$. Don´t forget to mark this answer as accepted. This is necessary since the totally wrong answer of Patricio has 3 upvotes. Very misleading. – callculus42 May 06 '21 at 17:00