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# Need some help with a easy task.

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Hi

I have a sum invested. Say 1000 dollars.

This sum will rise in value for 10 years with 10% per year. 1000*1.1^10. but I also want to add 100 dollars per year for 10 years and that amount will also rise in value by 10% per year.

Can someone plz help me with a complete formula for this? Thank you.

Guest Jan 25, 2015

#1
+91412
+5

This sum will rise in value for 10 years with 10% per year. 1000*1.1^10. but I also want to add 100 dollars per year for 10 years and that amount will also rise in value by 10% per year.

Can someone plz help me with a complete formula for this? Thank you.

Well if the 100 is put in at the beginning of each year for 10 years

The the first 100 will grow to 100*1.1^10

The the first 100 will grow to 100*1.1^9

The the last 100 will grow to 100*1.1^1

$$\\100*1.1^1+100*1.1^2+ .....100*1.1^{10}\\ =100(1.1+1.1^2+ ......+1.1^{10})\\\\ This is a GP a=100*1.1, r=1.1\\\\ sum = \frac{a(r^n-1)}{r-1}\\\\ sum = \frac{100*1.1(1.1^{10}-1)}{1.1-1}\\\\ sum = \frac{100*1.1(1.1^{10}-1)}{0.1}\\\\$$

So total after 10 years (all deposits are made at the beginning of the relevant year)

Grand Total = $$1000*1.1^{10}+ \frac{100*1.1(1.1^{10}-1)}{0.1}\\\\$$

Melody  Jan 26, 2015
Sort:

#1
+91412
+5

This sum will rise in value for 10 years with 10% per year. 1000*1.1^10. but I also want to add 100 dollars per year for 10 years and that amount will also rise in value by 10% per year.

Can someone plz help me with a complete formula for this? Thank you.

Well if the 100 is put in at the beginning of each year for 10 years

The the first 100 will grow to 100*1.1^10

The the first 100 will grow to 100*1.1^9

The the last 100 will grow to 100*1.1^1

$$\\100*1.1^1+100*1.1^2+ .....100*1.1^{10}\\ =100(1.1+1.1^2+ ......+1.1^{10})\\\\ This is a GP a=100*1.1, r=1.1\\\\ sum = \frac{a(r^n-1)}{r-1}\\\\ sum = \frac{100*1.1(1.1^{10}-1)}{1.1-1}\\\\ sum = \frac{100*1.1(1.1^{10}-1)}{0.1}\\\\$$

So total after 10 years (all deposits are made at the beginning of the relevant year)

Grand Total = $$1000*1.1^{10}+ \frac{100*1.1(1.1^{10}-1)}{0.1}\\\\$$

Melody  Jan 26, 2015

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