Your friend has invested in a 'Grow Your Money Scheme' that promises to return ₹11,000 after a year if you invest ₹1,000 at the rate of 10% compounded annually. You are a bit skeptic about the claim of this scheme. Your friend shows the calculation done by the agent of the scheme as follows. Would you be willing to invest in this scheme? Explain. [Scroll content — the agent's calculation:] Here is how I have done Given: $P = ₹1000$ $r = 10\%$ $n = 1$ year $A = P(1+r)^t$ $= 1000(1+10)^1$ $= 1000 \times 11$ $= ₹11{,}000$

> [!TLDR] The agent's calculation wrongly used $r=10$ instead of the decimal rate $r=0.10$; correctly, ₹1000 at 10% compounded annually for 1 year grows to only ₹1,100, not the claimed ₹11,000, so the scheme's claim is unreliable. > > [!FORMULA] $A=P(1+r)^t$ > $P$ = principal, $r$ = rate of interest per period expressed as a DECIMAL fraction (e.g. 10% = 0.10, not 10), $t$ = number of periods. 1. Given $P=₹1000$, stated annual rate $=10\%$, which as a decimal fraction is $r=0.10$ (NOT $r=10$), and $t=1$ year. 2. Correct substitution: $A=1000(1+0.10)^1=1000(1.10)$. 3. $1000\times1.10=₹1100$. 4. The agent's error: substituting $r=10$ (the bare percentage number) instead of $r=0.10$ gives $A=1000(1+10)^1=1000\times11=₹11{,}000$ — this silently treats "10% growth" as "1000% growth" (a factor of 11 instead of 1.1). 5. The genuine 10% compound growth on ₹1000 for 1 year is a gain of exactly ₹100 (i.e. $A=P+Pr=1000+100=1100$), matching step 3, not the agent's inflated ₹10,000 gain. > [!ANSWER] No — the correct amount after 1 year is ₹1,100 (10% of ₹1000 = ₹100 interest), not ₹11,000; the agent's formula wrongly used r=10 instead of r=0.10, so this scheme's claim is not credible as calculated.

Your friend has invested in a 'Grow Your Money Scheme' that promises to return ₹11,000 after a year if you inv

Exercise 5.4 · Q1

P = ₹1000

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✓ Free question

Concept understanding — Geometric Progression (GP)

Geometric Progression — From Intuition to Precision

Imagine you drop a ball from a height of 1 metre. It bounces back to half its previous height each time. The heights of successive bounces are:

$1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \dots$

What pattern do you see? Each term is obtained by multiplying the previous term by the same fixed number — here, $\frac{1}{2}$ . That is the core idea of a Geometric Progression (GP).

The Intuition

A GP is a sequence where the ratio between any two consecutive terms is constant. This constant is called the common ratio, denoted by $r$ .

If you start with a first term $a$ , the sequence looks like:

$a, \ ar, \ ar^2, \ ar^3, \ ar^4, \ \dots$

Each step multiplies by $r$ . If $r > 1$ , the terms grow (like population doubling). If $0 < r < 1$ , the terms shrink (like radioactive decay). If $r < 0$ , the terms alternate in sign.

Note

The word "geometric" comes from the idea that each term is the geometric mean of its neighbours — for three consecutive terms $x, y, z$ , we have $y^2 = xz$ .

The Precise Statement

A sequence $\{t_1, t_2, t_3, \dots\}$ is a Geometric Progression if there exists a constant $r$ such that for every $n \geq 1$ :

\frac{t_{n+1}}{t_n} = r

The first term is usually called $a$ , so:

t_n = a r^{\,n-1}

This is the $n$ th term formula. It lets you jump directly to any term without writing out the whole sequence.

t_n = a r^{\,n-1}

The Sum of a GP

Sometimes you need the sum of the first $n$ terms. Let $S_n = a + ar + ar^2 + \dots + ar^{\,n-1}$ .

Multiply the whole sum by $r$ :

r S_n = ar + ar^2 + ar^3 + \dots + ar^{\,n}

Subtract the second equation from the first:

S_n - r S_n = a - ar^{\,n}

S_n (1 - r) = a(1 - r^{\,n})

So:

S_n = \frac{a(1 - r^{\,n})}{1 - r}, \quad r \neq 1

If $r = 1$ , the sum is simply $S_n = n a$ .

Tip

For $|r| < 1$ , as $n \to \infty$ , $r^{\,n} \to 0$ , so the infinite sum converges to $\frac{a}{1 - r}$ . This is how recurring decimals like $0.\overline{3} = \frac{1}{3}$ are derived.

A Quick Example

Find the 10th term and the sum of the first 10 terms of the GP: $2, 6, 18, 54, \dots$

Here $a = 2$ , $r = \frac{6}{2} = 3$ .

10th term: $t_{10} = 2 \times 3^{9} = 2 \times 19683 = 39366$

Sum of first 10 terms:

S_{10} = \frac{2(3^{10} - 1)}{3 - 1} = \frac{2(59049 - 1)}{2} = 59048

Watch out

A common mistake: using $r^{\,n}$ instead of $r^{\,n-1}$ in the $n$ th term. The first term corresponds to $n=1$ , so the exponent is $n-1$ , not $n$ .

Why It Matters

GP appears everywhere — compound interest, population growth, depreciation, half-life in physics, and even in the geometry of fractals. Once you see the pattern of constant multiplication, you are looking at a geometric progression.

The agent's calculation wrongly used $r=10$ instead of the decimal rate $r=0.10$ ; correctly, ₹1000 at 10% compounded annually for 1 year grows to only ₹1,100, not the claimed ₹11,000, so the scheme's claim is unreliable.

[!FORMULA] $A=P(1+r)^t$

$P$ = principal, $r$ = rate of interest per period expressed as a DECIMAL fraction (e.g. 10% = 0.10, not 10), $t$ = number of periods.

Given $P=₹1000$ , stated annual rate $=10\%$ , which as a decimal fraction is $r=0.10$ (NOT $r=10$ ), and $t=1$ year.
Correct substitution: $A=1000(1+0.10)^1=1000(1.10)$ .
$1000\times1.10=₹1100$ .
The agent's error: substituting $r=10$ (the bare percentage number) instead of $r=0.10$ gives $A=1000(1+10)^1=1000\times11=₹11{,}000$ — this silently treats "10% growth" as "1000% growth" (a factor of 11 instead of 1.1).
The genuine 10% compound growth on ₹1000 for 1 year is a gain of exactly ₹100 (i.e. $A=P+Pr=1000+100=1100$ ), matching step 3, not the agent's inflated ₹10,000 gain.

✓Final answer

No — the correct amount after 1 year is ₹1,100 (10% of ₹1000 = ₹100 interest), not ₹11,000; the agent's formula wrongly used r=10 instead of r=0.10, so this scheme's claim is not credible as calculated.

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