Geometric Progression

A Constant Ratio

Look at $2, 4, 8, 16, \ldots$ — each term is twice the one before. In $\frac{1}{9}, -\frac{1}{27}, \frac{1}{81}, \ldots$ each term is $-\frac{1}{3}$ times the previous; in $.01, .0001, \ldots$ the multiplier is $.01$.

Key Point (Definition): A sequence $a_1, a_2, a_3, \ldots$ is a geometric progression (G.P.) if each term is non-zero and $\frac{a_{k+1}}{a_k} = r$ (constant) for all $k \ge 1$. Writing $a_1 = a$, the G.P. is $a, ar, ar^2, ar^3, \ldots$ — first term $a$, common ratio $r$.

The ratio may be negative (signs alternate) or fractional (terms shrink) — but no term is ever 0, and $r \ne 0$.

Key Point (General term): Each step multiplies by one more $r$, so

$$a_n = ar^{n-1}$$

— one $r$ fewer than the position: the 16th term is $ar^{15}$.

[Board Tip] Two-condition problems divide cleanly: 3rd term 24 and 6th term 192 give $\frac{ar^5}{ar^2} = r^3 = 8$, so $r = 2$, then $a = 6$. Divide the term equations — never subtract them.

Sum of the First n Terms

Multiply $S_n = a + ar + \cdots + ar^{n-1}$ by $r$ and subtract — the middle terms cancel:

Key Point (Sum formula):

$$S_n = \frac{a(r^n - 1)}{r - 1} = \frac{a(1 - r^n)}{1 - r}, \quad r \ne 1; \qquad S_n = na \text{ when } r = 1$$

For example: for $1 + \frac{2}{3} + \frac{4}{9} + \cdots$, $S_n = 3\left[1 - \left(\frac{2}{3}\right)^n\right]$ and $S_5 = \frac{211}{81}$. Run backwards: $3 + \frac{3}{2} + \frac{3}{4} + \cdots = \frac{3069}{512}$ forces $2^n = 1024$, i.e. $n = 10$ terms.

Almost-G.P.s: $7 + 77 + 777 + \cdots$ is not geometric, but $\frac{7}{9}(9 + 99 + 999 + \cdots)$ is a G.P. minus constants:

$$S_n = \frac{7}{9}\left[\frac{10(10^n - 1)}{9} - n\right]$$

[JEE Tip] The same three moves — factor the digit, write repunits as $10^k - 1$, sum the G.P. — handle $8 + 88 + \cdots$, $5 + 55 + \cdots$ and $.6 + .66 + \cdots$ as well. Learn the moves, not the answers.

Solved Examples

Example 1: Write the terms

Find the 10th and $n$-th terms of the G.P. $5, 25, 125, \ldots$

Solution:

Step 1 — Identify $a$ and $r$. $a = 5$; $r = \frac{25}{5} = 5$.

Step 2 — Apply $a_n = ar^{n-1}$. $a_{10} = 5 \cdot 5^9 = 5^{10}$.

Step 3 — General term. $a_n = 5 \cdot 5^{n-1} = 5^n$.

Takeaway: When $a = r$, the formula collapses to $a_n = r^n$ — worth noticing before computing.

Example 2: Which term is it?

Which term of the G.P. $2, 8, 32, \ldots$ is 131072?

Solution:

Step 1 — Set up the equation. $a = 2$, $r = 4$: $2 \cdot 4^{n-1} = 131072$.

Step 2 — Isolate the power. $4^{n-1} = 65536 = 4^8$.

Step 3 — Solve. $n - 1 = 8$, so $n = 9$: the 9th term.

Takeaway: "Which term" questions reduce to matching powers — write both sides with the same base.

Example 3: Two conditions

In a G.P. the 3rd term is 24 and the 6th term is 192. Find the 10th term.

Solution:

Step 1 — Write the two term equations. $ar^2 = 24$ and $ar^5 = 192$.

Step 2 — DIVIDE them. $\frac{ar^5}{ar^2} = r^3 = \frac{192}{24} = 8$, so $r = 2$.

Step 3 — Back-substitute. $a \cdot 4 = 24$ gives $a = 6$.

Step 4 — Compute the target. $a_{10} = 6 \cdot 2^9 = 6 \times 512 = 3072$.

Takeaway: Divide G.P. term equations (never subtract) — the $a$ cancels and only $r$ remains.

Example 4: Sum formula

Find $S_n$ and $S_5$ for the series $1 + \frac{2}{3} + \frac{4}{9} + \cdots$

Solution:

Step 1 — Identify. $a = 1$, $r = \frac{2}{3}$ ($|r| < 1$, so use the $1 - r^n$ form).

Step 2 — Apply the formula. $S_n = \frac{1 - (2/3)^n}{1 - 2/3} = 3\left[1 - \left(\frac{2}{3}\right)^n\right]$.

Step 3 — Evaluate $S_5$. $\left(\frac{2}{3}\right)^5 = \frac{32}{243}$: $S_5 = 3 \cdot \frac{211}{243} = \frac{211}{81}$.

Takeaway: Pick the version of the sum formula that keeps the bracket positive — fewer sign slips.

Example 5: Solve for n

How many terms of $3, \frac{3}{2}, \frac{3}{4}, \ldots$ give the sum $\frac{3069}{512}$?

Solution:

Step 1 — Write $S_n$. $a = 3$, $r = \frac{1}{2}$: $S_n = \frac{3(1 - 2^{-n})}{1/2} = 6\left(1 - \frac{1}{2^n}\right)$.

Step 2 — Equate and isolate. $6\left(1 - \frac{1}{2^n}\right) = \frac{3069}{512}$ gives $1 - \frac{1}{2^n} = \frac{3069}{3072}$.

Step 3 — Match powers. $\frac{1}{2^n} = \frac{3}{3072} = \frac{1}{1024} = \frac{1}{2^{10}}$: $n = 10$.

Takeaway: Solve-for-$n$ sums always end in a power match — keep everything as exact fractions.

Example 6: Sum and product of three terms

The first three terms of a G.P. sum to $\frac{13}{12}$ and their product is $-1$. Find the common ratio and the terms.

Solution:

Step 1 — Choose symmetric terms. Take $\frac{a}{r}, a, ar$: their product is $a^3$ — the $r$'s cancel.

Step 2 — Use the product. $a^3 = -1$ gives $a = -1$.

Step 3 — Use the sum. $-\frac{1}{r} - 1 - r = \frac{13}{12}$; multiply by $-12r$: $12 + 12r + 12r^2 = -13r$, i.e. $12r^2 + 25r + 12 = 0$.

Step 4 — Solve the quadratic. $r = \frac{-25 \pm 7}{24} = -\frac{3}{4}$ or $-\frac{4}{3}$.

Step 5 — Write the terms. $\frac{4}{3}, -1, \frac{3}{4}$ (or the same three reversed).

Takeaway: Symmetric choice $\frac{a}{r}, a, ar$ turns the product condition into $a^3$ instantly.

Example 7: The 7, 77, 777 sum

Find the sum of $7 + 77 + 777 + \cdots$ to $n$ terms.

Solution:

Step 1 — Factor out the digit. $S_n = 7(1 + 11 + 111 + \cdots) = \frac{7}{9}(9 + 99 + 999 + \cdots)$.

Step 2 — Write each repdigit as a power minus 1. $S_n = \frac{7}{9}\left[(10-1) + (10^2-1) + \cdots + (10^n-1)\right]$.

Step 3 — Sum the G.P. and the constants separately.

$$S_n = \frac{7}{9}\left[\frac{10(10^n - 1)}{9} - n\right]$$

Takeaway: Repdigit series are G.P.s in disguise — factor the digit, subtract the 1's.

Example 8: Ancestors

A person has 2 parents, 4 grandparents, 8 great-grandparents, and so on. Find the number of ancestors during the ten generations preceding his own.

Solution:

Step 1 — Recognise the G.P. Generations contribute $2, 4, 8, \ldots$: $a = 2$, $r = 2$, $n = 10$.

Step 2 — Sum. $S_{10} = \frac{2(2^{10} - 1)}{2 - 1} = 2(1024 - 1) = 2046$.

Takeaway: Real-world doubling processes are G.P.s — identify $a$, $r$, $n$ and the formula does the rest.

Example 9: From the 8th term back

Find the 12th term of a G.P. whose 8th term is 192 and common ratio is 2.

Solution:

Step 1 — Jump forward from the known term. $a_{12} = a_8 \cdot r^4$ — four more multiplications by $r$.

Step 2 — Evaluate. $192 \times 2^4 = 192 \times 16 = 3072$.

Step 3 — (Check via $a$.) $a \cdot 2^7 = 192$ gives $a = \frac{3}{2}$, and $\frac{3}{2} \cdot 2^{11} = 3072$ ✓.

Takeaway: $a_m = a_k \cdot r^{m-k}$ — you rarely need to find $a$ at all.

Example 10: A squared condition

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

Solution:

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

Step 2 — Cancel carefully. 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: G.P. terms are never zero, so dividing by powers of $a$ and $r$ is always legal.

Example 11: Surd ratio

Which term of $2, 2\sqrt{2}, 4, \ldots$ is 128?

Solution:

Step 1 — Identify the ratio. $r = \frac{2\sqrt{2}}{2} = \sqrt{2} = 2^{1/2}$.

Step 2 — Set up with base 2. $a_n = 2 \cdot 2^{\frac{n-1}{2}} = 2^{1 + \frac{n-1}{2}}$ and $128 = 2^7$.

Step 3 — Match exponents. $1 + \frac{n-1}{2} = 7$ gives $n - 1 = 12$: $n = 13$.

Takeaway: Surd ratios become half-integer exponents — the power-matching method is unchanged.

Example 12: G.P. condition

For what values of $x$ are $-\frac{2}{7}, x, -\frac{7}{2}$ in G.P.?

Solution:

Step 1 — Use the middle-term property. In a G.P., $x^2$ = product of the neighbours.

Step 2 — Compute. $x^2 = \left(-\frac{2}{7}\right)\left(-\frac{7}{2}\right) = 1$.

Step 3 — Solve. $x = \pm 1$ — both give a consistent (negative) common ratio.

Takeaway: Three-in-G.P. means middle$^2$ = ends' product; expect two signs unless context kills one.

Example 13: A sigma with two parts

Evaluate $\sum_{k=1}^{11} (2 + 3^k)$.

Solution:

Step 1 — Split the sum. $\sum_{k=1}^{11} 2 + \sum_{k=1}^{11} 3^k = 22 + (3 + 3^2 + \cdots + 3^{11})$.

Step 2 — Sum the G.P. $\frac{3(3^{11} - 1)}{3 - 1} = \frac{3(177147 - 1)}{2} = \frac{531438}{2} = 265719$.

Step 3 — Add. $22 + 265719 = 265741$.

Takeaway: Split mixed sigmas into a constant part and a G.P. part — each has its own one-line formula.

Example 14: How many terms

How many terms of the G.P. $3, 3^2, 3^3, \ldots$ are needed to give the sum 120?

Solution:

Step 1 — Write $S_n$. $a = 3$, $r = 3$: $S_n = \frac{3(3^n - 1)}{2}$.

Step 2 — Equate. $\frac{3(3^n - 1)}{2} = 120$ gives $3^n - 1 = 80$, so $3^n = 81 = 3^4$.

Step 3 — Conclude and check. $n = 4$: $3 + 9 + 27 + 81 = 120$ ✓.

Takeaway: After matching the power, spend five seconds adding the actual terms — a free full check.

Example 15: Two blocks of three

The sum of the first three terms of a G.P. is 16 and the sum of the next three is 128. Find $a$, $r$ and $S_n$.

Solution:

Step 1 — Relate the blocks. Terms 4-6 are terms 1-3 each multiplied by $r^3$: block ratio $r^3 = \frac{128}{16} = 8$.

Step 2 — Solve for $r$. $r = 2$.

Step 3 — Find $a$. $a(1 + r + r^2) = a \cdot 7 = 16$ gives $a = \frac{16}{7}$.

Step 4 — Write $S_n$. $S_n = \frac{16}{7}(2^n - 1)$.

Takeaway: Consecutive equal-length blocks of a G.P. are in ratio $r^{\text{(block length)}}$ — one division finds $r$.

Example 16: $S_7$ from $a$ and $a_7$

Given a G.P. with $a = 729$ and 7th term 64, determine $S_7$.

Solution:

Step 1 — Find $r$. $729\,r^6 = 64$ gives $r^6 = \frac{64}{729} = \left(\frac{2}{3}\right)^6$; take $r = \frac{2}{3}$.

Step 2 — Apply the sum formula. $S_7 = \frac{729\left(1 - (2/3)^7\right)}{1 - 2/3} = 3 \cdot 729\left(1 - \frac{128}{2187}\right)$.

Step 3 — Evaluate. $2187 - 2187 \cdot \frac{128}{2187} = 2187 - 128 = 2059$.

Takeaway: $729 = 3^6$ and $64 = 2^6$ — recognising perfect powers turns $r^6$ into a one-look solve.