Solved Examples

How to Use This Section

Thirty fully worked examples sweep the whole chapter — general-term and which-term drills, G.P.s pinned down by two conditions, sum computations forwards and backwards, the sum-with-product classics, repdigit series, growth applications from bacteria to chain letters, G.M. insertions, A.M.-G.M. recoveries, and the classic proof-style questions — instalment plans, the S-P-R identity and the dwindling workforce — solved in full.

[Board Tip] For any three-term G.P. problem, choose the terms $\frac{a}{r}, a, ar$ — the product collapses to $a^3$ instantly. For four terms use $\frac{a}{r^3}, \frac{a}{r}, ar, ar^3$ (ratio $r^2$) when the product is given. The symmetric choice is half the solution.

Terms and Conditions

Example 1: Fractional G.P.

Find the 20th and $n$-th terms of the G.P. $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \ldots$

Solution:

Step 1 — Identify. $a = \frac{5}{2}$, $r = \frac{1}{2}$.

Step 2 — General term. $a_n = \frac{5}{2}\left(\frac{1}{2}\right)^{n-1} = \frac{5}{2^n}$.

Step 3 — Evaluate. $a_{20} = \frac{5}{2^{20}}$.

Takeaway: Fold the first term's denominator into the power — $\frac{5}{2^n}$ is cleaner than the raw formula.

Example 2: Three terms in G.P. within a G.P.

The 5th, 8th and 11th terms of a G.P. are $p, q, s$. Show $q^2 = ps$.

Solution:

Step 1 — Write the three terms. $p = ar^4$, $q = ar^7$, $s = ar^{10}$.

Step 2 — Multiply the outer pair. $ps = a^2r^{14}$.

Step 3 — Compare with the middle. $q^2 = (ar^7)^2 = a^2r^{14} = ps$. ∎

Takeaway: Equally spaced terms of a G.P. are themselves in G.P. — the spacing (3 here) only changes the new ratio.

Example 3: Square condition

The 4th term of a G.P. is the square of its 2nd term and the first term is $-3$. Determine the 7th term.

Solution:

Step 1 — Translate. $ar^3 = (ar)^2 = a^2r^2$.

Step 2 — Cancel. Divide by $ar^2$ (non-zero): $r = a = -3$.

Step 3 — Compute. $a_7 = ar^6 = -3 \cdot (-3)^6 = -3 \times 729 = -2187$.

Takeaway: $(-3)^6$ is positive — the lone leading $-3$ carries the final sign.

Example 4: Which term

(i) Which term of $\sqrt{3}, 3, 3\sqrt{3}, \ldots$ is 729? (ii) Which term of $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots$ is $\frac{1}{19683}$?

Solution:

Step 1 — (i) Base-3 exponents. $a = 3^{1/2}$, $r = 3^{1/2}$: $a_n = 3^{n/2}$. Set $3^{n/2} = 729 = 3^6$: $n = 12$.

Step 2 — (ii) Matching reciprocals. $a_n = \left(\frac{1}{3}\right)^n = \frac{1}{19683} = \frac{1}{3^9}$: $n = 9$.

Takeaway: Everything to one base, then equate exponents — surd ratios just produce half-integer exponents.

Example 5: Equal spacing again

If the 4th, 10th and 16th terms of a G.P. are $x, y, z$, prove $x, y, z$ are in G.P.

Solution:

Step 1 — Write the terms. $x = ar^3$, $y = ar^9$, $z = ar^{15}$.

Step 2 — Form both ratios. $\frac{y}{x} = r^6$ and $\frac{z}{y} = r^6$.

Step 3 — Conclude. Equal ratios: $x, y, z$ is a G.P. with common ratio $r^6$. ∎

Takeaway: The gap (6 positions) becomes the exponent of the derived ratio — same principle as Example 2.

Example 6: Four numbers

Find four numbers in G.P. whose third term is greater than the first by 9, and whose second term is greater than the fourth by 18.

Solution:

Step 1 — Translate both conditions. $ar^2 - a = 9$ and $ar - ar^3 = 18$, i.e. $a(r^2 - 1) = 9$ and $-ar(r^2 - 1) = 18$.

Step 2 — Divide. $\frac{-ar(r^2-1)}{a(r^2-1)} = -r = \frac{18}{9} = 2$, so $r = -2$.

Step 3 — Back-substitute. $a(4 - 1) = 9$ gives $a = 3$.

Step 4 — Write and check. $3, -6, 12, -24$: third exceeds first by $12 - 3 = 9$ ✓; second exceeds fourth by $-6 - (-24) = 18$ ✓.

Takeaway: Dividing paired conditions cancels the shared factor $a(r^2 - 1)$ — the sign of the quotient carries the answer.

Sums

Example 7: Decimal G.P.

Find the sum of $0.15, 0.015, 0.0015, \ldots$ to 20 terms.

Solution:

Step 1 — Identify. $a = 0.15$, $r = 0.1$.

Step 2 — Apply the formula. $S_{20} = \frac{0.15\left(1 - (0.1)^{20}\right)}{1 - 0.1} = \frac{0.15}{0.9}\left[1 - (0.1)^{20}\right]$.

Step 3 — Simplify. $\frac{0.15}{0.9} = \frac{1}{6}$: $S_{20} = \frac{1}{6}\left[1 - (0.1)^{20}\right]$.

Takeaway: Leave tiny powers like $(0.1)^{20}$ symbolic — the exact form is the expected answer.

Example 8: Alternating powers

Find the sum of $1, -a, a^2, -a^3, \ldots$ to $n$ terms ($a \ne -1$).

Solution:

Step 1 — Identify the ratio. Each term is $-a$ times the previous: $r = -a$.

Step 2 — Apply the formula. $S_n = \frac{1 - (-a)^n}{1 - (-a)} = \frac{1 - (-a)^n}{1 + a}$.

Takeaway: Keep $(-a)^n$ intact — its sign depends on the parity of $n$, and the formula handles both at once.

Example 9: Odd powers

Find the sum of $x^3, x^5, x^7, \ldots$ to $n$ terms ($x \ne \pm 1$).

Solution:

Step 1 — Identify. $a = x^3$, $r = x^2$.

Step 2 — Apply. $S_n = \frac{x^3\left((x^2)^n - 1\right)}{x^2 - 1} = \frac{x^3\left(x^{2n} - 1\right)}{x^2 - 1}$.

Takeaway: The restriction $x \ne \pm 1$ is exactly $r \ne 1$ — the formula's only demand.

Example 10: Sum and product, product 1

The first three terms of a G.P. sum to $\frac{39}{10}$ and their product is 1. Find the common ratio and the terms.

Solution:

Step 1 — Symmetric terms. $\frac{a}{r}, a, ar$: product $= a^3 = 1$, so $a = 1$.

Step 2 — Sum condition. $\frac{1}{r} + 1 + r = \frac{39}{10}$, so $r + \frac{1}{r} = \frac{29}{10}$.

Step 3 — Solve. $10r^2 - 29r + 10 = 0$ gives $r = \frac{29 \pm 21}{20} = \frac{5}{2}$ or $\frac{2}{5}$.

Step 4 — Write the terms. $\frac{2}{5}, 1, \frac{5}{2}$ (both ratios give the same three numbers).

Takeaway: Reciprocal ratio pairs describe the same G.P. read in opposite directions.

Example 11: Sum of two terms plus a ratio condition

Find a G.P. whose first two terms sum to $-4$ and whose fifth term is 4 times the third.

Solution:

Step 1 — Ratio condition first. $ar^4 = 4ar^2$ gives $r^2 = 4$: $r = \pm 2$.

Step 2 — Case $r = 2$. $a(1 + 2) = -4$: $a = -\frac{4}{3}$ — G.P. $-\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \ldots$

Step 3 — Case $r = -2$. $a(1 - 2) = -4$: $a = 4$ — G.P. $4, -8, 16, -32, \ldots$

Takeaway: Both sign cases survive here — report both G.P.s; nothing in the problem eliminates either.

Example 12: Repdigit sum

Find the sum of $8, 88, 888, \ldots$ to $n$ terms.

Solution:

Step 1 — Factor the digit. $S_n = \frac{8}{9}(9 + 99 + 999 + \cdots)$.

Step 2 — Powers minus ones. $= \frac{8}{9}\left[(10-1) + (10^2-1) + \cdots + (10^n - 1)\right]$.

Step 3 — Sum both parts. $S_n = \frac{8}{9}\left[\frac{10(10^n - 1)}{9} - n\right]$.

Takeaway: Identical to the 7-77-777 recipe — only the leading digit changes.

Example 13: Products of paired terms

Find the sum of the products of corresponding terms of $2, 4, 8, 16, 32$ and $128, 32, 8, 2, \frac{1}{2}$.

Solution:

Step 1 — Form the products. $256, 128, 64, 32, 16$.

Step 2 — Recognise the G.P. $a = 256$, $r = \frac{1}{2}$ (product of the two ratios $2 \times \frac{1}{4}$).

Step 3 — Sum. $S_5 = \frac{256\left(1 - \frac{1}{32}\right)}{\frac{1}{2}} = 512 \times \frac{31}{32} = 496$.

Takeaway: Products of paired G.P. terms form a G.P. with ratio $r_1 r_2$ — the next example proves it in general.

Example 14: Products in general

Show that the products of corresponding terms of $a, ar, \ldots, ar^{n-1}$ and $A, AR, \ldots, AR^{n-1}$ form a G.P., and find its common ratio.

Solution:

Step 1 — Write the $k$-th product. $ar^{k-1} \cdot AR^{k-1} = aA\,(rR)^{k-1}$.

Step 2 — Read off the structure. First term $aA$, each step multiplies by $rR$ — a G.P. with common ratio $rR$. ∎

Takeaway: One-line proofs fall out when you write the general ($k$-th) term rather than listing cases.

Growth and Money

Example 15: Bacteria

Bacteria double every hour, starting from 30. How many are present at the end of the 2nd, 4th and $n$-th hours?

Solution:

Step 1 — Model the doubling. Each hour multiplies by 2: after $n$ hours, $30 \cdot 2^n$.

Step 2 — Evaluate. 2nd hour: $30 \times 4 = 120$; 4th hour: $30 \times 16 = 480$.

Step 3 — General. $n$-th hour: $30 \cdot 2^n$.

Takeaway: "End of the $n$-th hour" means $n$ doublings — the count is $a r^n$, not $ar^{n-1}$, because the start is hour 0.

Example 16: Compound interest

What will Rs 500 amount to in 10 years at 10% per annum compounded annually?

Solution:

Step 1 — One year's growth. Adding 10% multiplies by $1.1$.

Step 2 — Ten years. Amount $= 500(1.1)^{10}$.

Step 3 — Evaluate. $(1.1)^{10} \approx 2.5937$: amount $\approx$ Rs $1296.87$.

Takeaway: Compound interest IS a G.P. — the year-end amounts $500(1.1), 500(1.1)^2, \ldots$ have ratio $1.1$.

Example 17: Depreciation

A machine costing Rs 15625 depreciates 20% each year. Find its value after 5 years.

Solution:

Step 1 — One year's loss. Losing 20% multiplies by $0.8$.

Step 2 — Five years. Value $= 15625(0.8)^5$.

Step 3 — Evaluate exactly. $15625 = 5^6$ and $(0.8)^5 = \frac{4^5}{5^5} = \frac{1024}{3125}$: value $= 5 \times 1024 = 5120$ rupees.

Takeaway: Depreciation is the same G.P. machinery with $r < 1$ — and $15625 = 5^6$ makes the arithmetic exact.

Example 18: Chain letters

Each of 4 friends mails the letter to 4 more, and so on. At 50 paise per letter, find the postage when the 8th set is mailed.

Solution:

Step 1 — Count the letters. Rounds mail $4, 4^2, \ldots, 4^8$ letters: $S_8 = \frac{4(4^8 - 1)}{3}$.

Step 2 — Evaluate. $4^8 = 65536$: $S_8 = \frac{4 \times 65535}{3} = 87380$ letters.

Step 3 — Price it. $87380 \times \frac{1}{2} =$ Rs $43690$.

Takeaway: Chain-letter growth is a G.P. sum, not a single term — every round's letters cost postage.

Example 19: Simple interest contrast

Rs 10000 is deposited at 5% simple interest. Find the amount in the 15th year and the total after 20 years.

Solution:

Step 1 — Recognise the A.P. Simple interest adds a FLAT Rs 500 per year — arithmetic, not geometric growth.

Step 2 — 15th year. After 14 completed years: $10000 + 14 \times 500 = 17000$.

Step 3 — After 20 years. $10000 + 20 \times 500 = 20000$.

Takeaway: Simple interest → A.P.; compound interest → G.P. — the contrast is the whole point of this pair.

Example 20: The tractor

A farmer buys a tractor for Rs 12000, pays Rs 6000 cash, and clears the balance in annual instalments of Rs 500 plus 12% interest on the unpaid amount. Find the total cost.

Solution:

Step 1 — List the unpaid balances. $6000, 5500, 5000, \ldots, 500$ — an A.P. with 12 instalments.

Step 2 — Total interest. $0.12 \times \left(6000 + 5500 + \cdots + 500\right) = 0.12 \times \frac{12}{2}(6000 + 500) = 0.12 \times 39000 = 4680$.

Step 3 — Total cost. $12000 + 4680 =$ Rs $16680$.

Takeaway: Interest on a shrinking balance sums over an A.P. — pair the first and last balances and multiply.

Example 21: The scooter

Shamshad Ali buys a scooter for Rs 22000, pays Rs 4000 cash, and clears the balance in instalments of Rs 1000 plus 10% interest on the unpaid amount. Find the total cost.

Solution:

Step 1 — Unpaid balances. $18000, 17000, \ldots, 1000$ — 18 instalments.

Step 2 — Interest. $0.10 \times \frac{18}{2}(18000 + 1000) = 0.10 \times 171000 = 17100$.

Step 3 — Total cost. $22000 + 17100 =$ Rs $39100$.

Takeaway: Same structure as the tractor — only the numbers change; learn the frame, not the figures.

Miscellaneous Classics

Example 22: A functional equation

$f$ satisfies $f(x+y) = f(x)f(y)$ for all $x, y \in \mathbb{N}$, with $f(1) = 3$ and $\sum_{x=1}^{n} f(x) = 120$. Find $n$.

Solution:

Step 1 — Generate the values. $f(2) = f(1)^2 = 9$, $f(3) = 27$, and inductively $f(x) = 3^x$.

Step 2 — Sum the G.P. $3 + 9 + \cdots + 3^n = \frac{3(3^n - 1)}{2} = 120$.

Step 3 — Solve. $3^n - 1 = 80$, so $3^n = 81$: $n = 4$.

Takeaway: $f(x+y) = f(x)f(y)$ forces exponential values — the functional equation IS the G.P. definition.

Example 23: Last term from the sum

The sum of some terms of a G.P. is 315; the first term is 5 and the ratio 2. Find the last term and the number of terms.

Solution:

Step 1 — Set up. $S_n = 5(2^n - 1) = 315$.

Step 2 — Solve for $n$. $2^n - 1 = 63$, so $2^n = 64$: $n = 6$.

Step 3 — Last term. $a_6 = 5 \cdot 2^5 = 160$.

Takeaway: Sum questions often hide a power match — then the last term is one substitution away.

Example 24: Ratio from a sum of terms

The first term of a G.P. is 1 and the sum of its third and fifth terms is 90. Find the common ratio.

Solution:

Step 1 — Translate. $r^2 + r^4 = 90$.

Step 2 — Substitute $t = r^2$. $t^2 + t - 90 = 0$ gives $t = 9$ or $t = -10$.

Step 3 — Reject and finish. $t = r^2 \ge 0$ kills $-10$: $r^2 = 9$, so $r = \pm 3$.

Takeaway: Even-power substitutions create phantom negative roots — audit them against $t = r^2 \ge 0$.

Example 25: G.P. meets A.P.

Three numbers in G.P. sum to 56. Subtracting $1, 7, 21$ from them in order gives an A.P. Find the numbers.

Solution:

Step 1 — Set up both conditions. $a(1 + r + r^2) = 56$; A.P. means $2(ar - 7) = (a - 1) + (ar^2 - 21)$.

Step 2 — Simplify the A.P. condition. $2ar - 14 = a + ar^2 - 22$, so $a(r - 1)^2 = 8$.

Step 3 — Divide the conditions. $\frac{1 + r + r^2}{(r-1)^2} = 7$ gives $2r^2 - 5r + 2 = 0$: $r = 2$ or $\frac{1}{2}$.

Step 4 — Finish. $r = 2$: $a \cdot 7 = 56$, $a = 8$. Numbers: $8, 16, 32$. Check: $7, 9, 11$ is an A.P. ✓

Takeaway: Mixed G.P./A.P. conditions: simplify each separately, then divide to kill $a$.

Example 26: Odd-place terms

A G.P. has an even number of terms. If the sum of all terms is 5 times the sum of the terms in odd places, find the common ratio.

Solution:

Step 1 — Describe the odd-place terms. With $2n$ terms, positions $1, 3, 5, \ldots$ hold $a, ar^2, ar^4, \ldots$ — a G.P. with ratio $r^2$ and $n$ terms.

Step 2 — Form both sums. $S_{\text{all}} = \frac{a(r^{2n} - 1)}{r - 1}$; $S_{\text{odd}} = \frac{a(r^{2n} - 1)}{r^2 - 1}$.

Step 3 — Divide. $\frac{S_{\text{all}}}{S_{\text{odd}}} = \frac{r^2 - 1}{r - 1} = r + 1 = 5$: $r = 4$.

Takeaway: The unknown-length sums cancel entirely — only $r + 1$ survives the division.

Example 27: S, P, R identity

For $n$ terms of a G.P., let $S$ be the sum, $P$ the product and $R$ the sum of reciprocals. Prove $P^2R^n = S^n$.

Solution:

Step 1 — Write the three quantities. $S = \frac{a(r^n - 1)}{r - 1}$; the reciprocals form a G.P. with first term $\frac{1}{a}$ and ratio $\frac{1}{r}$: $R = \frac{r^n - 1}{ar^{n-1}(r - 1)}$; and $P = a^n r^{0+1+\cdots+(n-1)} = a^n r^{\frac{n(n-1)}{2}}$.

Step 2 — Form $\frac{S}{R}$. The brackets cancel: $\frac{S}{R} = a^2 r^{n-1}$.

Step 3 — Raise to the $n$. $\left(\frac{S}{R}\right)^n = a^{2n} r^{n(n-1)} = \left(a^n r^{\frac{n(n-1)}{2}}\right)^2 = P^2$, i.e. $P^2R^n = S^n$. ∎

Takeaway: Never expand $P^2$ first — form $\frac{S}{R}$ and the messy sums vanish.

Example 28: Two quadratics

$a, b$ are roots of $x^2 - 3x + p = 0$; $c, d$ are roots of $x^2 - 12x + q = 0$; and $a, b, c, d$ form a G.P. Prove $(q + p) : (q - p) = 17 : 15$.

Solution:

Step 1 — Name the G.P. $a, ar, ar^2, ar^3$. Root sums: $a(1 + r) = 3$ and $ar^2(1 + r) = 12$.

Step 2 — Divide. $r^2 = 4$.

Step 3 — Form $p$ and $q$. $p = a \cdot ar = a^2r$; $q = ar^2 \cdot ar^3 = a^2r^5$.

Step 4 — Take the ratio. $\frac{q + p}{q - p} = \frac{a^2r(r^4 + 1)}{a^2r(r^4 - 1)} = \frac{16 + 1}{16 - 1} = \frac{17}{15}$. ∎

Takeaway: Products of root-pairs of a G.P. differ by a power of $r$ — everything else cancels in the ratio.

Example 29: 5 + 55 + 555

Find the sum of $5 + 55 + 555 + \cdots$ to $n$ terms.

Solution:

Step 1 — Factor the digit. $S_n = \frac{5}{9}(9 + 99 + \cdots) = \frac{5}{9}\left[(10 - 1) + (10^2 - 1) + \cdots + (10^n - 1)\right]$.

Step 2 — Sum. $S_n = \frac{5}{9}\left[\frac{10(10^n - 1)}{9} - n\right]$.

Takeaway: Third appearance of the repdigit recipe — by now it should take under a minute.

Example 30: The dwindling workforce

150 workers start a job; 4 drop out each subsequent day. The job takes 8 more days than planned. In how many days was the work completed?

Solution:

Step 1 — Express the planned work. With all 150 staying, the planned duration $n$ gives $150n$ worker-days.

Step 2 — Express the actual work. Workforce $150, 146, 142, \ldots$ for $n + 8$ days: $\sum_{k=0}^{n+7}(150 - 4k) = 150(n+8) - 4 \cdot \frac{(n+7)(n+8)}{2}$.

Step 3 — Equate and simplify. $150(n+8) - 2(n+7)(n+8) = 150n$ gives $150 \times 8 = 2(n+7)(n+8)$, so $(n+7)(n+8) = 600$.

Step 4 — Factor. $600 = 24 \times 25$: $n = 17$; the work took $n + 8 = 25$ days.

Takeaway: Match consecutive-integer products by factoring the target — $24 \times 25 = 600$ ends the problem.