help with compound payments
- Thread starter Sue0113
- Start date
reworked problem
800=PMT[1-(1.0852)^-.25]
----------------
.0852
800=PMT[1-(1.889154807)
800=PMTx.889154807
PMT=800/.889154807
PMT=899.7308384
899.73/4=224.93
PMT=224.93
So in order to pay the friend back he would have to make 4 payments of 224.93. Is this correct?
800=PMT[1-(1.0852)^-.25]
----------------
.0852
800=PMT[1-(1.889154807)
800=PMTx.889154807
PMT=800/.889154807
PMT=899.7308384
899.73/4=224.93
PMT=224.93
So in order to pay the friend back he would have to make 4 payments of 224.93. Is this correct?
Hello, Sue0113!
If you're careful, you can "walk" your way through this one.
But wear your hiking boots, it's a long walk.
Let \(\displaystyle P\) = the principal, $800.
Let \(\displaystyle i\) = the interest rate, \(\displaystyle \frac{0.0852}{4} = 0.0213\)
Let \(\displaystyle X\) = the quarterly payment
Note: If he owes \(\displaystyle D\) dollars at the beginning of a quarter,
. . . . at the end of the quarter, he will owe: \(\displaystyle D + Di \,=\,D(1+i)\) dollars.
At the beginning of the year, he owes you \(\displaystyle P\) dollars.
End of 1st quarter:
With interest, he owes: .\(\displaystyle P(1+i)\) dollars.
He repays \(\displaystyle X\) dollars.
His balance is: .\(\displaystyle P(1+i) - X\) dollars.
End of 2nd quarter:
With interest, he owes:. \(\displaystyle \big[P(1+i) - X\big](1+i)\)
. . \(\displaystyle =\
(1+i)^2 -X(1+i)\) dollars.
He repays \(\displaystyle X\) dollars.
His balance is: .\(\displaystyle P(1+i)^2 - X(1+i) - X\) dollars.
End of 3rd quarter:
With interest, he owes: .\(\displaystyle \big[P(1+i)^2 - X(1+i) - X\big](1+i)\)
. . \(\displaystyle =\(1+i)^3 - X(1+i)^2 - X(1+i)\) dollars.
He repays \(\displaystyle X\) dollars.
His balance is: .\(\displaystyle P(1+i)^3 - X(1+i)^2 - X(1+i) - X\) dollars.
End of 4th quarter:
With interest, he owes: .\(\displaystyle \big[P(1+i)^3 - X(1+i)^2 - X(1+i) - X\big](1+i)\)
. . \(\displaystyle -\(1+i)^4 - X(1+i)^3 - X(1+i)^2 - X(1+i)\) dollars.
He repays \(\displaystyle X\) dollars.
His balance is: .\(\displaystyle P(1+i)^4 - X(1+i)^3 - X(1+i)^2 - X(1+i) - X\) dollars.
But at this time, he has completey repaid the loan; his balance is zero.
. . \(\displaystyle P(1+i)^4 - X(1+i)^3 - X(1+i)^2 - X(1+i) - X \;=\;0\)
. . \(\displaystyle P(1 + i)^4 \;=\;X(1+i)^3 + X(1+i)^2 + X(1+i) + X\)
. . \(\displaystyle P(1+i)^4 \;=\;X\big[(1+i)^3 + (1+i)^2 + (1+i) + 1\big]\)
. . \(\displaystyle P(1+i)^4 \;=\;X\,\dfrac{(1+i)^4-1}{(1+i) - 1}\)
. . \(\displaystyle P(1+i)^4 \;=\;X\,\dfrac{(1+i)^4-1}{i}\)
\(\displaystyle \text{Therefore: }\;X \;=\;P\,\dfrac{i(1+i)^4}{(1+i)^4-1} \)
We have: .\(\displaystyle P = 800,\;i = 0.0213\)
\(\displaystyle \text{Therefore: }\:X \;=\;800\,\dfrac{(0.0213)(1.0213)^4}{(1.0213^4-1} \;=\; 210.7622178\)
His payments are $210.76 per quarter.
If you're careful, you can "walk" your way through this one.
But wear your hiking boots, it's a long walk.
Let \(\displaystyle P\) = the principal, $800.
Let \(\displaystyle i\) = the interest rate, \(\displaystyle \frac{0.0852}{4} = 0.0213\)
Let \(\displaystyle X\) = the quarterly payment
Note: If he owes \(\displaystyle D\) dollars at the beginning of a quarter,
. . . . at the end of the quarter, he will owe: \(\displaystyle D + Di \,=\,D(1+i)\) dollars.
At the beginning of the year, he owes you \(\displaystyle P\) dollars.
End of 1st quarter:
With interest, he owes: .\(\displaystyle P(1+i)\) dollars.
He repays \(\displaystyle X\) dollars.
His balance is: .\(\displaystyle P(1+i) - X\) dollars.
End of 2nd quarter:
With interest, he owes:. \(\displaystyle \big[P(1+i) - X\big](1+i)\)
. . \(\displaystyle =\
(1+i)^2 -X(1+i)\) dollars.He repays \(\displaystyle X\) dollars.
His balance is: .\(\displaystyle P(1+i)^2 - X(1+i) - X\) dollars.
End of 3rd quarter:
With interest, he owes: .\(\displaystyle \big[P(1+i)^2 - X(1+i) - X\big](1+i)\)
. . \(\displaystyle =\
(1+i)^3 - X(1+i)^2 - X(1+i)\) dollars.He repays \(\displaystyle X\) dollars.
His balance is: .\(\displaystyle P(1+i)^3 - X(1+i)^2 - X(1+i) - X\) dollars.
End of 4th quarter:
With interest, he owes: .\(\displaystyle \big[P(1+i)^3 - X(1+i)^2 - X(1+i) - X\big](1+i)\)
. . \(\displaystyle -\
(1+i)^4 - X(1+i)^3 - X(1+i)^2 - X(1+i)\) dollars.He repays \(\displaystyle X\) dollars.
His balance is: .\(\displaystyle P(1+i)^4 - X(1+i)^3 - X(1+i)^2 - X(1+i) - X\) dollars.
But at this time, he has completey repaid the loan; his balance is zero.
. . \(\displaystyle P(1+i)^4 - X(1+i)^3 - X(1+i)^2 - X(1+i) - X \;=\;0\)
. . \(\displaystyle P(1 + i)^4 \;=\;X(1+i)^3 + X(1+i)^2 + X(1+i) + X\)
. . \(\displaystyle P(1+i)^4 \;=\;X\big[(1+i)^3 + (1+i)^2 + (1+i) + 1\big]\)
. . \(\displaystyle P(1+i)^4 \;=\;X\,\dfrac{(1+i)^4-1}{(1+i) - 1}\)
. . \(\displaystyle P(1+i)^4 \;=\;X\,\dfrac{(1+i)^4-1}{i}\)
\(\displaystyle \text{Therefore: }\;X \;=\;P\,\dfrac{i(1+i)^4}{(1+i)^4-1} \)
We have: .\(\displaystyle P = 800,\;i = 0.0213\)
\(\displaystyle \text{Therefore: }\:X \;=\;800\,\dfrac{(0.0213)(1.0213)^4}{(1.0213^4-1} \;=\; 210.7622178\)
His payments are $210.76 per quarter.
We have not learnt Annuity
Denis we are not into annuity as of yet however we may get into it.
after reading both the explainations I now have a further understand but Soroban explained it as we are learning
FV=x+E4+E3+E2+E1
So your both right and you both helped me understand.
Denis we are not into annuity as of yet however we may get into it.
after reading both the explainations I now have a further understand but Soroban explained it as we are learning
FV=x+E4+E3+E2+E1
So your both right and you both helped me understand.