Worked examples · Sheet 3

Problems with geometric progressions

Theory used: Geometric progression Sum of the terms of a geometric progression Limit of the geometric sequence

Exercise 1

In a geometric progression with positive terms we have $a_2 = 6$ and $a_5 = 162$ . Find the common ratio, the first term, the eighth term and the sum of the first eight terms.

\frac{a_5}{a_2} = q^3 \iff q^3 = \frac{162}{6} = 27 \iff q = 3

a_1 = \frac{a_2}{q} = 2, \qquad a_8 = a_1\,q^{7} = 2 \cdot 2187 = 4374

S_8 = a_1\,\frac{1 - q^{8}}{1 - q} = 2 \cdot \frac{1 - 6561}{1 - 3} = 6560

In a geometric progression, to go from $a_2$ to $a_5$ one multiplies by the common ratio three times, so the ratio $\dfrac{a_5}{a_2}$ equals $q^3$ . From $q^3 = 27$ we get $q = 3$ , the only real cube root. The first term is obtained by dividing $a_2$ by the common ratio, and the eighth with the formula $a_n = a_1\,q^{n-1}$ . The sum of the terms is computed with the formula for the geometric sum; as a check, $2 + 6 + 18 + 54 + 162 + 486 + 1458 + 4374 = 6560$ .

Exercise 2

At the end of every month, $200$ euros are deposited, for $24$ months, into an account that yields $0.5\%$ per month with compound interest. Compute the balance immediately after the last deposit.

S = 200 + 200 \cdot 1.005 + 200 \cdot 1.005^{2} + \dots + 200 \cdot 1.005^{23}

S = 200 \cdot \frac{1.005^{24} - 1}{1.005 - 1} \approx 200 \cdot 25.4320 \approx 5086.39 \text{ euros}

Each deposit stays in the account for a different number of months. The last one stays in the account zero months, the second-to-last one month and the first $23$ months, because it is made at the end of the first month. A deposit that stays $k$ months is multiplied by $1.005^{k}$ , so the balance is the sum of $24$ terms of a geometric progression with first term $200$ and common ratio $1.005$ . With the formula for the sum of the terms, written as $a_1\,\dfrac{q^n - 1}{q - 1}$ so that numerator and denominator are positive, the balance is about $5086.39$ euros. The deposits total $4800$ euros, so the interest amounts to about $286$ euros.

Exercise 3

A machine bought for $20\,000$ euros loses $15\%$ of its value every year. Find after how many years its value falls below $5000$ euros.

V_n = 20\,000 \cdot 0.85^{\,n}

20\,000 \cdot 0.85^{\,n} < 5000 \iff 0.85^{\,n} < 0.25 \iff n > \frac{\ln 0.25}{\ln 0.85} \approx 8.53

V_8 \approx 5449.81, \qquad V_9 \approx 4632.34

Losing $15\%$ means keeping $85\%$ , so every year the value is multiplied by $0.85$ and the values form a geometric progression with common ratio $0.85 < 1$ , which tends to $0$ by the limit of the geometric sequence. In the inequality one takes the logarithm of both sides; then one divides by $\ln 0.85$ , which is negative, and the direction of the inequality is reversed. This gives $n > 8.53$ , and the first natural number that satisfies the condition is $9$ . Therefore the value falls below $5000$ euros after 9 years, as confirmed by $V_8$ and $V_9$ .