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

Question

Find the sum of the GP: $2, 6, 18, 54, ...$ up to 8 terms. Also find the sum to infinity of: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ...$

Solution — Step by Step

Here $a = 2$ , $r = 6/2 = 3$ , $n = 8$ . Since $|r| > 1$ :

S_n = a \cdot \frac{r^n - 1}{r - 1} = 2 \cdot \frac{3^8 - 1}{3 - 1} = 2 \cdot \frac{6561 - 1}{2} = \mathbf{6560}

Here $a = 1$ , $r = 1/2$ . Since $|r| < 1$ , the infinite sum converges:

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

Why This Works

graph TD
    A["GP Sum: Which formula?"] --> B["|r| > 1?"]
    B -->|Yes| C["S_n = a times r^n - 1 over r - 1"]
    B -->|No| D["|r| < 1?"]
    D -->|Yes| E["S_n = a times 1 - r^n over 1 - r"]
    D --> F["Want infinite sum?"]
    F -->|"Yes, and |r| < 1"| G["S_∞ = a / 1 - r"]
    F -->|"|r| ≥ 1"| H["Infinite sum DIVERGES"]
    A --> I["r = 1?"]
    I -->|Yes| J["S_n = na, all terms equal"]

The finite GP sum formula comes from multiplying $S_n$ by $r$ and subtracting: $S_n - rS_n = a - ar^n$ , giving $S_n(1-r) = a(1-r^n)$ .

For infinite GP with $|r| < 1$ : as $n \to \infty$ , $r^n \to 0$ , so $S_\infty = a/(1-r)$ . The terms keep getting smaller fast enough that the total stays bounded. This is why $1 + 1/2 + 1/4 + ...$ adds to exactly 2, even though there are infinitely many terms.

Compound interest connection: If you invest Rs $P$ at rate $r$ per year compounded annually, after $n$ years you have $P(1+r)^n$ . This is the $n$ th term of a GP with first term $P$ and common ratio $(1+r)$ . The total amount deposited in an SIP (investing Rs $P$ every year) is a GP sum: $S_n = P \cdot \frac{(1+r)^n - 1}{r}$ .

Alternative Method

A quick check: for the infinite GP $1 + 1/2 + 1/4 + ...$ , you can verify by partial sums. $S_1 = 1$ , $S_2 = 1.5$ , $S_3 = 1.75$ , $S_4 = 1.875$ . The sums approach 2 but never exceed it. This gives intuition for why the answer is 2.

For JEE problems: if a repeating decimal like $0.333...$ appears, recognise it as an infinite GP: $3/10 + 3/100 + 3/1000 + ... = (3/10)/(1 - 1/10) = 3/9 = 1/3$ . This is how we prove that $0.\overline{3} = 1/3$ .

Common Mistake

Applying the infinite GP formula when $|r| \geq 1$ . The formula $S_\infty = a/(1-r)$ works ONLY when $|r| < 1$ . If $|r| \geq 1$ , the series diverges (the terms do not shrink, so the sum grows without bound). Students sometimes blindly apply the formula to get a finite answer — but that answer is meaningless. Always check $|r| < 1$ before using the infinite sum formula.

$n$ th term: $a_n = ar^{n-1}$

Sum of $n$ terms: $S_n = a \cdot \frac{r^n - 1}{r - 1}$ (when $r \neq 1$ )

Sum to infinity ( $|r| < 1$ ): $S_\infty = \frac{a}{1 - r}$

Geometric mean of $a$ and $b$ : $G = \sqrt{ab}$

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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