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

How many terms of G.P. 3, 3², 3³, … are needed to give the sum 120?

Exercise 8.2 · Q13

3, 3^2, 3^3, \ldots

Manipur CohsemTextbookSubjective· 2mImportance★★★★★est

24% · 27/114 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Geometric Progression

Geometric Progression: The Idea of Repeated Multiplication

Imagine you're folding a piece of paper in half. Start with thickness 1 unit. After one fold, thickness becomes 2. After two folds, thickness becomes 4. After three folds, thickness becomes 8. The sequence of thicknesses is:

1, 2, 4, 8, 16, ...

Notice the pattern: each term is obtained by multiplying the previous term by the same number (here, 2). That's the core intuition behind a geometric progression — you keep multiplying by a fixed number, step after step.

This is different from an arithmetic progression, where you keep adding a fixed number. Here, the growth is multiplicative, not additive. That's why geometric progressions grow (or shrink) much faster.

Precise Definition

A Geometric Progression (GP) is a sequence of numbers where the ratio of any term to its preceding term is constant. This constant is called the common ratio, denoted by $r$ .

If the first term is $a$ , then the sequence looks like:

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

Note

The common ratio $r$ can be any real number — positive, negative, or even a fraction. If $r$ is negative, the terms alternate in sign. If $0 < r < 1$ , the terms get smaller and smaller.

The $n$ -th Term

To find any term directly without listing all previous ones, use the formula:

T_n = a \cdot r^{\,n-1}

where $T_n$ is the $n$ -th term, $a$ is the first term, $r$ is the common ratio, and $n$ is the term number (starting from 1).

Example: For the paper-folding sequence, $a = 1$ , $r = 2$ . The 5th term is $1 \cdot 2^{5-1} = 2^4 = 16$ , which matches our list.

Sum of $n$ Terms

There are two cases, depending on whether $r = 1$ or not.

n

When $r = 1$ , every term is just $a$ , so the sum is simply $n \times a$ .

Why the formula works (intuition):

Let $S = a + ar + ar^2 + \dots + ar^{n-1}$ . Multiply both sides by $r$ : $rS = ar + ar^2 + \dots + ar^n$ . Subtract the first from the second: $rS - S = ar^n - a$ , so $S(r-1) = a(r^n - 1)$ , giving the formula above.

Sum of an Infinite GP

If the common ratio $r$ lies strictly between $-1$ and $1$ (i.e., $|r| < 1$ ), the terms get smaller and smaller, and the sum of all terms approaches a finite value:

S_\infty = \frac{a}{1 - r}, \quad \text{for } |r| < 1

Watch out

If $|r| \geq 1$ , the infinite sum does not exist (it diverges to infinity or oscillates without settling). Never apply the infinite sum formula when $|r| \geq 1$ .

Example: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$ has $a=1$ , $r=\frac12$ , so $S_\infty = \frac{1}{1 - 1/2} = 2$ . This matches the intuition that repeatedly halving a unit length eventually fills exactly 2 units.

Quick Reference Table

Property	Formula	Condition
Common ratio	$r = \dfrac{T_{n+1}}{T_n}$	Always
$n$ -th term	$T_n = a r^{n-1}$	Always

Start your 14-day free trial to unlock the full solution →

This is a geometric progression with first term $a = 3$ and common ratio $r = 3$ . Using the sum formula $S_n = a\frac{r^n - 1}{r - 1}$ , we set $S_n = 120$ and solve for $n$ , obtaining $n = 4$ . So 4 terms are needed.

The problem asks: how many terms of the GP $3, 3^2, 3^3, \ldots$ must be added to get a total of 120?

A geometric progression grows by multiplying each term by a fixed ratio. Here, the first term is $3$ , and each subsequent term is $3$ times the previous one — so the ratio is $3$ . The terms are $3, 9, 27, 81, 243, \ldots$

If you add the first few: $3 + 9 = 12$ , $12 + 27 = 39$ , $39 + 81 = 120$ . That’s exactly four terms. But let’s confirm it algebraically, because exam questions often require the formula approach.

n

Here $a = 3$ , $r = 3$ , and we want $S_n = 120$ .

Set up the equation

120 = 3 \cdot \frac{3^n - 1}{3 - 1} = 3 \cdot \frac{3^n - 1}{2}

Simplify Multiply both sides by $2$ :

240 = 3(3^n - 1)

Divide by $3$ :

80 = 3^n - 1

Solve for $n$ $3^n = 81$ …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026Set ANNUAL1 markMCQ
$Q. A person has two parents, four grandparents, eight great grandparents and so on. Then the number of his ancestors during the ten generations preceeding to his own is$
(a) 1084
(b) 1024
(c) 2250
(d) 2046
›Reveal solutionSolution

Summing the geometric series 2+4+8+...+2^10 (10 generations) gives 2046 ancestors.

The number of ancestors in generation k is $2^k$ (2 parents, 4 grandparents, 8 great-grandparents, ...). Over 10 generations, the total is the geometric series:

…

Start your 14-day free trial to see the full solution →
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026Set ANNUAL1 mark
$x+9, x-6, 4$
›Reveal solutionSolution

Using the G.P. condition (middle term)² = product of the outer terms, x = 0 or x = 16.

For x+9, x−6, 4 to be in G.P., $(x-6)^2 = (x+9)(4)$ .

Expand: $x^2-12x+36 = 4x+36$

$x^2-12x-4x = 0$

$x^2-16x = 0$

$x(x-16)=0$

So $x=0$ or $x=16$ .

…

Start your 14-day free trial to see the full solution →
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2025Set ANNUAL1 markMCQ
$1, \dfrac{1}{3}, \dfrac{1}{9}, \ldots$
(a) Infinity
(b) 0
(c) $\dfrac{1}{3}$
(d) $\dfrac{3}{2}$
›Reveal solutionSolution

For the G.P. $1, \tfrac13, \tfrac19,\ldots$ with $a=1$ , $r=\tfrac13$ , the infinite sum is $S_\infty = \dfrac{a}{1-r} = \dfrac{3}{2}$ .

Here $a = 1$ and the common ratio $r = \dfrac{1/3}{1} = \dfrac13$ . Since $|r| = \tfrac13 < 1$ , the series converges and …

Start your 14-day free trial to see the full solution →
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2025Set ANNUAL1 markMCQ
$x = \pm 1$
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is false.
›Reveal solutionSolution

Applying $b^2=ac$ to $-\tfrac27, x, -\tfrac72$ gives $x^2=1$ , i.e. $x=\pm1$ , exactly matching the Assertion — so both statements are true and (R) explains (A).

For three numbers $a,b,c$ to be in G.P., the middle term squared must equal the product of the outer terms: $b^2 = ac$ — this is exactly Reason (R), which is a true, standard fact about G.P.s.

Apply it to $-\dfrac27,\ x,\ -\dfrac72$ : treating $x$ as the middle term,

$x^2 = \left(-\dfrac27\right)\left(-\dfrac72\right) = 1 \implies x = \pm 1.$

…

Start your 14-day free trial to see the full solution →
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2023Set ANNUAL1 markMCQ
$\dfrac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}$
(a) $\dfrac{1}{2}$
(b) $-\dfrac{1}{2}$
(c) 1
(d) -1
›Reveal solutionSolution

The value of $n$ that makes $(a^{n+1}+b^{n+1})/(a^n+b^n)$ equal to the geometric mean $\sqrt{ab}$ of $a$ and $b$ is $n=-\dfrac12$ .

The G.M. between $a$ and $b$ is $\sqrt{ab}$ . We are told

$\dfrac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}} = \sqrt{ab}$

Step 1: Try $n=-\dfrac12$ .

$\dfrac{a^{1/2}+b^{1/2}}{a^{-1/2}+b^{-1/2}}$

Multiply numerator and denominator by $\sqrt{ab}$ :

$\dfrac{\sqrt{ab}(a^{1/2}+b^{1/2})}{\sqrt{ab}\,a^{-1/2} + \sqrt{ab}\,b^{-1/2}} = \dfrac{\sqrt{ab}(\sqrt a+\sqrt b)}{\sqrt b + \sqrt a}$

The factor $(\sqrt a+\sqrt b)$ cancels top and bottom, leaving exactly $\sqrt{ab}$ — confirming $n=-\dfrac12$ works. …

Start your 14-day free trial to see the full solution →
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2022Set ANNUAL1 mark
$x$
›Reveal solutionSolution

Using $x^2 = \left(-\dfrac27\right)\left(-\dfrac72\right)$ gives $x^2=1$ , so $x=\pm1$ .

Three numbers $a,b,c$ are in G.P. exactly when $b^2=ac$ (the middle term is the geometric mean of the outer two).

Here $a=-\dfrac27$ , $b=x$ , $c=-\dfrac72$ , so:

…

Start your 14-day free trial to see the full solution →

←Previous question · 8.5Q12 — The sum of first three terms of a G.P. is (39)/(10) and their product …→Next question · 8.5Q14 — The sum of first three terms of a G.P. is 16 and the sum of the next t…

Q.How many terms of G.P. 3,32,33,…3, 3^2, 3^3, \ldots3,32,33,… are needed to give the sum 120?

🎓Unlock everything free for 14 days

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

Unlock everything free for 14 days