a)
90−65=2590-65=2590−65=25
P=S∗i∗(1+i)n(1+i)n−1P=S*\frac{i*(1+i)^n}{(1+i)^n-1}P=S∗(1+i)n−1i∗(1+i)n
25,000=S∗0.08∗1.08251.0825−125,000=S*\frac{0.08*1.08^{25}}{1.08^{25}-1}25,000=S∗1.0825−10.08∗1.0825
S=266,808.96S=266,808.96S=266,808.96
b)
65−32=3365-32=3365−32=33
S∗=b1(qn+1−1)q−1−b1S^*=\frac{b1(q^{n+1}-1)}{q-1}-b1S∗=q−1b1(qn+1−1)−b1
S∗=315,253.34S^*=315,253.34S∗=315,253.34
S∗>SS^*>SS∗>S
Yes it is enough
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