Simple and Compound Interest, Loans: Problems with Solutions

Solution:
Each year the same part of the principal is repaid:
[tex]\frac{26000}{4}=6500[/tex]
Hence at the beginning of each of the four years the outstanding debt is [tex]26000[/tex], [tex]19500[/tex], [tex]13000[/tex] and [tex]6500[/tex] dollars, respectively. The interest for each year is [tex]3.6\%[/tex] of the debt at the beginning of that year:
[tex]26000\times 0.036=936, \quad 19500\times 0.036=702, \quad 13000\times 0.036=468, \quad 6500\times 0.036=234[/tex]
The total interest is
[tex]936+702+468+234=2340[/tex]
So [tex]2340[/tex] dollars of interest are paid in total.

Solution:
With annual compound interest of [tex]6\%[/tex] a sum grows [tex]1.06[/tex] times every year. The first income stays in the bank for [tex]2[/tex] years:
[tex]50000\times 1.06^2=50000\times 1.1236=56180[/tex]
The second income stays in the bank for [tex]1[/tex] year:
[tex]50000\times 1.06=53000[/tex]
The third income is deposited at the end of the third year and earns no interest. In total:
[tex]56180+53000+150000=259180[/tex]
So at the end of the third year there will be [tex]259180[/tex] dollars in the bank. For comparison, if the [tex]190000[/tex] dollars were deposited at [tex]6\%[/tex] instead of being invested, after three years they would become only [tex]190000\times 1.06^3=226293.04[/tex] dollars, so the investment yields more than [tex]6\%[/tex] per year.

Solution:
Let [tex]q=1.028[/tex] and let [tex]r[/tex] be the yearly installment. At the end of each year the debt is multiplied by [tex]q[/tex] and then [tex]r[/tex] is paid, so after the first year, the second year, and so on up to the fourth year, the debt is, respectively,
[tex]37000q-r, \quad 37000q^2-rq-r, \quad\ldots, \quad 37000q^4-r\left(q^3+q^2+q+1\right)[/tex]
After the fourth year the loan is repaid, so the last expression equals [tex]0[/tex]. Since [tex]q^3+q^2+q+1=\frac{q^4-1}{q-1}[/tex], we get
[tex]r=\frac{37000\times q^4(q-1)}{q^4-1}=\frac{37000\times 1.028^4\times 0.028}{1.028^4-1}\approx 9906.44[/tex]
So each installment is approximately [tex]9906.44[/tex] dollars.

Solution:
First count the deposits: from March to December [tex]2005[/tex] there are [tex]10[/tex] of them, in the years from [tex]2006[/tex] to [tex]2015[/tex] there are [tex]10\times 12=120[/tex], and from January to April [tex]2016[/tex] there are [tex]4[/tex]. In total:
[tex]10+120+4=134[/tex]
Let [tex]q=1.004[/tex]. Each deposit is made at the beginning of a month and earns interest for that month as well. So the deposit of April [tex]1[/tex], [tex]2016[/tex] grows to [tex]500q[/tex], the deposit of March [tex]1[/tex], [tex]2016[/tex] grows to [tex]500q^2[/tex], and so on, up to the first deposit, which earns interest for [tex]134[/tex] months and grows to [tex]500q^{134}[/tex]. The sum of this geometric progression is
[tex]500\left(q+q^2+\ldots+q^{134}\right)=500q\times \frac{q^{134}-1}{q-1}=\frac{500\times 1.004}{0.004}\left(1.004^{134}-1\right)=125500\left(1.004^{134}-1\right)[/tex]
So the account holds [tex]125500\left(1.004^{134}-1\right)\approx 88769.94[/tex] dollars.

Solution:
Each month the balance is multiplied by [tex]q[/tex]. Saving phase: there are [tex]40\times 12=480[/tex] deposits. The deposit made at the end of month [tex]k[/tex] earns interest for [tex]480-k[/tex] months, so right after the last deposit the account holds
[tex]S=100\left(q^{479}+q^{478}+\ldots+q+1\right)=\frac{100\left(q^{480}-1\right)}{q-1}[/tex]
Payout phase: there are [tex]15\times 12=180[/tex] payments of [tex]P[/tex] dollars. After [tex]n[/tex] months the balance is [tex]Sq^n-P\left(q^{n-1}+\ldots+q+1\right)[/tex], and after [tex]180[/tex] months it must be [tex]0[/tex]:
[tex]Sq^{180}=P\left(q^{179}+\ldots+q+1\right)=P\times \frac{q^{180}-1}{q-1}[/tex]
Hence
[tex]P=\frac{Sq^{180}(q-1)}{q^{180}-1}=\frac{100\times q^{180}\left(q^{480}-1\right)}{q^{180}-1}\approx 1206.77[/tex]
So the monthly pension is approximately [tex]1206.77[/tex] dollars (the accumulated sum is [tex]S\approx 152602.02[/tex] dollars).