GP properties — sum to n terms, infinite GP, applications in compound interest

Question

Find the sum of the first 8 terms of the GP: $3, 6, 12, 24, \ldots$ . Also, find the sum to infinity of the GP: $1, 1/2, 1/4, 1/8, \ldots$ . How does GP relate to compound interest?

(CBSE 11 & JEE Main — sequences and series)

Solution — Step by Step

First GP: $a = 3$ , $r = 6/3 = 2$ , $n = 8$ .

Since $|r| > 1$ , use:

S_n = a \cdot \frac{r^n - 1}{r - 1} = 3 \cdot \frac{2^8 - 1}{2 - 1} = 3 \times 255 = \mathbf{765}

Second GP: $a = 1$ , $r = 1/2$ .

Since $|r| < 1$ , the infinite sum converges:

S_\infty = \frac{a}{1 - r} = \frac{1}{1 - 1/2} = \frac{1}{1/2} = \mathbf{2}

The partial sums ( $1, 1.5, 1.75, 1.875, \ldots$ ) get closer and closer to $2$ but never exceed it.

Compound interest creates a GP. If principal $= P$ and rate $= r\%$ per year:

Year 1: $P(1 + r/100)$ Year 2: $P(1 + r/100)^2$ Year $n$ : $P(1 + r/100)^n$

This is a GP with first term $P(1 + r/100)$ and common ratio $(1 + r/100)$ .

The total amount after $n$ years in an SIP (equal annual deposits of $P$ ) is the sum of this GP.

Why This Works

In a GP, each term is a fixed multiple ( $r$ ) of the previous one. The sum formula exploits the telescoping property: multiply $S_n$ by $r$ and subtract from $S_n$ — most terms cancel.

graph TD
    A["GP Problem"] --> B{"What's asked?"}
    B -->|"Sum of n terms"| C{"r = 1?"}
    C -->|"Yes"| D["S = na"]
    C -->|"No"| E["S = a(rⁿ - 1)/(r - 1)"]
    B -->|"Sum to infinity"| F{"|r| < 1?"}
    F -->|"Yes"| G["S∞ = a/(1 - r)"]
    F -->|"No"| H["Diverges — no finite sum"]
    B -->|"nth term"| I["aₙ = a × r^(n-1)"]
    A --> J{"Application?"}
    J -->|"Compound interest"| K["Amount = P(1+r/100)ⁿ"]
    J -->|"Recurring decimals"| L["0.333... = 3/10 + 3/100 +...<br/>= (3/10)/(1-1/10) = 1/3"]

The infinite sum exists only when $|r| < 1$ because the terms get smaller and smaller, adding less each time. If $|r| \geq 1$ , the terms don’t shrink (or grow), so the sum diverges.

Alternative Method — Recurring Decimals as Infinite GPs

$0.999\ldots = 9/10 + 9/100 + 9/1000 + \ldots$

This is a GP with $a = 9/10$ , $r = 1/10$ :

S_\infty = \frac{9/10}{1 - 1/10} = \frac{9/10}{9/10} = 1

So $0.999\ldots = 1$ exactly. This elegant proof uses the infinite GP formula.

For JEE: the GM-AM inequality states that for positive numbers, the geometric mean never exceeds the arithmetic mean: $\sqrt{ab} \leq (a+b)/2$ . Combined with properties of GPs, this inequality solves many optimisation problems. If three terms $a, b, c$ are in GP, then $b^2 = ac$ .

Common Mistake

Students apply the infinite sum formula $S_\infty = a/(1-r)$ when $|r| \geq 1$ . The formula is valid ONLY when the common ratio has absolute value strictly less than 1. For $r = 2$ , the sum $1 + 2 + 4 + 8 + \ldots$ diverges to infinity — there is no finite answer. Always check $|r| < 1$ before using this formula.

Question

Solution — Step by Step

Why This Works

Alternative Method — Recurring Decimals as Infinite GPs

Common Mistake

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