Exercise 6.5 — nth Term of GP

General (nth) term of a geometric progression.

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Exercise 6.5 — Geometric Progressions (Finding the nth Term)

Exercise 6.5 is part of Chapter 6, Progressions, in Class 10 Mathematics for students following the CBSE, Telangana, and Andhra Pradesh board syllabi. This exercise focuses entirely on Geometric Progressions (GP) — sequences in which every term is obtained by multiplying the previous term by a fixed number called the common ratio.

Unlike an Arithmetic Progression (AP), where consecutive terms differ by a constant added amount, a Geometric Progression grows or shrinks by a constant multiplying factor. This exercise builds your skill in identifying the common ratio, writing the general nth term formula, and solving real exam-style problems involving unknown terms, unknown ratios, and two GPs with a common term.

Common Ratio (r) nth Term Formula Finding Specific Terms Forming a GP from Given Terms Equal nth Terms of Two GPs
CORE FORMULA
The nth term of a Geometric Progression with first term a and common ratio r is given by: aₙ = a·r^(n−1)
💡 Why this formula matters: Every single question in this exercise — from finding r, to finding a specific term, to comparing two GPs — uses this one formula: aₙ = a·rn−1. Master this, and the entire exercise becomes a matter of substitution and algebra.

How the nth Term Formula Is Derived

Before solving problems, it helps to see where the formula aₙ = a·rn−1 actually comes from. Consider the general form of a GP: a, ar, ar², ar³, ar⁴, … Each term is the previous term multiplied by the common ratio r.

a₁ = a
a₂ = ar
a₃ = ar²
a₄ = ar³
a₅ = ar⁴

Notice the pattern in the exponents: the 1st term has exponent 0, the 2nd term has exponent 1, the 3rd term has exponent 2, and so on. In general, the exponent is always one less than the term number. This pattern generalizes to:

aₙ = a · r^(n−1)

This single formula lets you calculate any term of a GP — whether it's the 5th term or the 50th term — as long as you know the first term a and the common ratio r.

Question 1 — Find the Common Ratio r and the nth Term aₙ

In this question, you're given four different geometric progressions and asked to find the common ratio r for each, then write the general nth term formula. The common ratio is always found by dividing any term by the term immediately before it: r = a₂ ÷ a₁.

Part (i)
G.P.: 3, 3/2, 3/4, 3/8, ...
First term: a = 3 Common ratio: r = a₂ ÷ a₁ = (3/2) ÷ 3 = 1/2 nth term: aₙ = a·r^(n−1) aₙ = 3 × (1/2)^(n−1) = 3 / 2^(n−1)
r
1/2
Common ratio
aₙ
3/2ⁿ⁻¹
General term
Part (ii)
G.P.: 2, −6, 18, −54, ...
First term: a = 2 Common ratio: r = a₂ ÷ a₁ = (−6) ÷ 2 = −3 nth term: aₙ = a·r^(n−1) aₙ = 2 × (−3)^(n−1)
r
−3
Common ratio
aₙ
2(−3)ⁿ⁻¹
General term
📌 Note: A negative common ratio makes the signs of the terms alternate — positive, negative, positive, negative, and so on. This is a useful pattern check when verifying your answer.
Part (iii)
G.P.: −1, −3, −9, −27, ...
First term: a = −1 Common ratio: r = a₂ ÷ a₁ = (−3) ÷ (−1) = 3 nth term: aₙ = a·r^(n−1) aₙ = (−1) × 3^(n−1)
r
3
Common ratio
aₙ
−1·3ⁿ⁻¹
General term
Part (iv)
G.P.: 5, 2, 4/5, 8/25, ...
First term: a = 5 Common ratio: r = a₂ ÷ a₁ = 2 ÷ 5 = 2/5 nth term: aₙ = a·r^(n−1) aₙ = 5 × (2/5)^(n−1)
r
2/5
Common ratio
aₙ
5(2/5)ⁿ⁻¹
General term

Quick Answer Summary — Question 1

PartG.P.raₙ
(i)3, 3/2, 3/4, 3/8, ...1/23/2ⁿ⁻¹
(ii)2, −6, 18, −54, ...−32(−3)ⁿ⁻¹
(iii)−1, −3, −9, −27, ...3−1(3)ⁿ⁻¹
(iv)5, 2, 4/5, 8/25, ...2/55(2/5)ⁿ⁻¹

Question 2 — Find the 10th and nth Term of the G.P. 5, 25, 125, ...

This question asks for two things: the specific 10th term and the general nth term formula. Both require finding the first term and common ratio first.

Solution
Find a₁₀ and aₙ for the G.P. 5, 25, 125, ...
First term: a = 5 Common ratio: r = a₂ ÷ a₁ = 25 ÷ 5 = 5 Finding a₁₀: aₙ = a·r^(n−1) ⟹ a₁₀ = a·r⁹ a₁₀ = 5 × 5⁹ = 5¹⁰ a₁₀ = 5¹⁰ Finding aₙ: aₙ = a·r^(n−1) = 5 × 5^(n−1) aₙ = 5ⁿ
Shortcut spotted: Since a = 5 = 5¹ and r = 5, multiplying 5¹ by 5^(n−1) simply adds the exponents: 5^(1+n−1) = 5ⁿ. Recognizing when the first term itself equals the common ratio can simplify your final answer significantly.

Question 3 — Find the Indicated Term of Each Geometric Progression

Here, you are directly given the first term a and the common ratio r, and asked to find a specific term using the nth term formula.

Part (i)
a₁ = 9, r = 1/3, find a₇
a₇ = a·r⁶ = 9 × (1/3)⁶ 9 = 3² , so: a₇ = 3² × (1/3)⁶ = 3² × 1/3⁶ = 1/3^(6−2) = 1/3⁴ a₇ = 1/81
Part (ii)
a₁ = −12, r = 1/3, find a₆
a₆ = a·r⁵ = (−12) × (1/3)⁵ = −12 × 1/243 = −4 × 1/81 (simplifying −12/243) a₆ = −4/81
Part (i)
1/81
a₇, when a=9, r=1/3
Part (ii)
−4/81
a₆, when a=−12, r=1/3
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Question 4 — Which Term of the G.P. Equals a Given Value?

This is a reverse problem: instead of finding a term's value, you're given the value and must find which position (n) it occupies in the sequence. The strategy is to set aₙ equal to the given value, substitute the formula, and solve for n by matching exponents.

Part (i)
Which term of 2, 8, 32, ... is 512?
First term: a = 2; Common ratio: r = 8 ÷ 2 = 4 Let the nth term be 512: aₙ = a·r^(n−1) = 512 2 × 4^(n−1) = 512 4^(n−1) = 512 ÷ 2 = 256 4^(n−1) = 4⁴ (since 256 = 4⁴) ⟹ n − 1 = 4 ∴ n = 5 (the 5th term is 512)
Part (ii)
Which term of 3, 3√3, 9, ... is 729?
First term: a = 3; Common ratio: r = √3 Let the nth term be 729: aₙ = a·r^(n−1) = 729 3 × (√3)^(n−1) = 729 3 × 3^((n−1)/2) = 3⁶ (since 729 = 3⁶) 3^(1 + (n−1)/2) = 3⁶ ⟹ n/2 = 6 ∴ n = 12 (the 12th term is 729)
Part (iii)
Which term of 1/3, 1/9, 1/27, ... is 1/2187?
First term: a = 1/3; Common ratio: r = 1/3 Let the nth term be 1/2187: aₙ = a·r^(n−1) = 1/2187 (1/3) × (1/3)^(n−1) = 1/2187 (1/3)ⁿ = 1/2187 = 1/3⁷ (since 2187 = 3⁷) ∴ n = 7 (the 7th term is 1/2187)
💡 Problem-solving pattern: Whenever you're asked "which term equals a given value," always express both sides as powers of the same base. Once the bases match, simply equate the exponents to solve for n.

Question 5 — Find the 12th Term Using the 8th Term and Common Ratio

Given that the 8th term of a GP is 192 and the common ratio is 2, find the 12th term. Instead of finding the first term separately, this question is solved more efficiently using the relationship between terms that are a fixed number of positions apart.

Solution
Find a₁₂, given a₈ = 192 and r = 2
a₁₂ = a·r^(12−1) = a·r¹¹ Rewrite: a·r¹¹ = (a·r⁷) × r⁴ = a₈ × r⁴ (splitting r¹¹ = r⁷ × r⁴) a₁₂ = 192 × 2⁴ a₁₂ = 192 × 16 a₁₂ = 3072
Efficient technique: Since the 12th term is exactly 4 positions after the 8th term, you can jump directly from a₈ to a₁₂ by multiplying by r⁴ — no need to calculate the first term 'a' at all. This shortcut works whenever you know any single term and the common ratio.
aₘ = aₖ × r^(m−k) (jumping from the kth term to the mth term)

Question 6 — Find the GP When Two Terms Are Given

The 4th term of a GP is 2/3, and the 7th term is 16/81. This question asks you to find the complete geometric series — meaning you must determine both the first term a and the common ratio r from two equations.

Solution
Find the G.P. when a₄ = 2/3 and a₇ = 16/81
Set up two equations: a₄ = a·r³ = 2/3 …(1) a₇ = a·r⁶ = 16/81 …(2) Divide equation (2) by equation (1): (a·r⁶) ÷ (a·r³) = (16/81) ÷ (2/3) r³ = (16/81) × (3/2) = 8/27 r³ = (2/3)³ ⟹ r = 2/3 Substitute r = 2/3 into equation (1): a × (2/3)³ = 2/3 a × 8/27 = 2/3 a = (2/3) × (27/8) = 9/4 ∴ a = 9/4 and r = 2/3
The required G.P. is: 9/4, (9/4)(2/3), (9/4)(2/3)², (9/4)(2/3)³, … which simplifies to: 9/4, 3/2, 1, 2/3, …
📌 General method for "find the GP" problems: When two non-consecutive terms are given, divide the equation for the higher-indexed term by the equation for the lower-indexed term. This eliminates 'a' and leaves an equation purely in r, which can then be solved and substituted back.

Question 7 — Two GPs With Equal nth Terms

This is one of the more advanced questions in the exercise. Two separate geometric progressions, 162, 54, 18, ... and 2/81, 2/27, 2/9, ..., are given. We are told their nth terms are equal for some value of n, and we must find that value of n.

Solution
Find n when the nth terms of both G.P.s are equal
First G.P.: 162, 54, 18, ... a = 162; r = 54 ÷ 162 = 1/3 Second G.P.: 2/81, 2/27, 2/9, ... a = 2/81; r = (2/27) ÷ (2/81) = 3 Set the nth terms equal: 162 × (1/3)^(n−1) = (2/81) × 3^(n−1) 162 × (81/2) = 3^(n−1) × 3^(n−1) 6561 = 3^(2(n−1)) 3⁸ = 3^(2n−2) (since 6561 = 3⁸) ⟹ 2n − 2 = 8 ⟹ 2n = 10 ∴ n = 5
Verification tip: You can verify this by computing the 5th term of each GP separately. The 5th term of the first GP is 162 × (1/3)⁴ = 2, and the 5th term of the second GP is (2/81) × 3⁴ = 2. Both equal 2 — confirming n = 5 is correct.

Common Mistakes to Avoid

  • Computing r incorrectly: Always divide a later term by the term immediately before it (a₂ ÷ a₁), never the reverse. Dividing a₁ ÷ a₂ gives the reciprocal of r and leads to wrong answers throughout the problem.
  • Forgetting the (n−1) exponent: A very common error is writing aₙ = a·rⁿ instead of the correct aₙ = a·r^(n−1). Always double-check the exponent is one less than the term number.
  • Sign errors with negative common ratios: When r is negative (as in Question 1(ii) and (iii)), be careful with parentheses — (−3)^(n−1) is very different from −3^(n−1). Always keep the negative sign and the base together in brackets.
  • Mixing up "jump" formula direction: In problems like Question 5, the relation aₘ = aₖ × r^(m−k) only works when m is greater than k. If asked to go backward (e.g., from a₁₂ to a₈), you would divide by r^(m−k) instead of multiplying.
  • Not simplifying powers to a common base: In "which term is this value" problems (Question 4), always express the given value as a power of the same base as the common ratio. Skipping this step makes the equation impossible to solve directly.
Exam trap: In Question 6 and Question 7 type problems, students often forget to substitute the value of r back into one of the original equations to find 'a'. Finding r alone is not a complete answer — you must report both a and r (or the full series) as required by the question.

Quick Reference — All Answers at a Glance

QuestionGivenAnswer
Q1(i)3, 3/2, 3/4, 3/8, ...r = 1/2, aₙ = 3/2ⁿ⁻¹
Q1(ii)2, −6, 18, −54, ...r = −3, aₙ = 2(−3)ⁿ⁻¹
Q1(iii)−1, −3, −9, −27, ...r = 3, aₙ = −1(3)ⁿ⁻¹
Q1(iv)5, 2, 4/5, 8/25, ...r = 2/5, aₙ = 5(2/5)ⁿ⁻¹
Q25, 25, 125, ...a₁₀ = 5¹⁰, aₙ = 5ⁿ
Q3(i)a=9, r=1/3, find a₇a₇ = 1/81
Q3(ii)a=−12, r=1/3, find a₆a₆ = −4/81
Q4(i)2, 8, 32, ... = 512?5th term
Q4(ii)3, 3√3, 9, ... = 729?12th term
Q4(iii)1/3, 1/9, 1/27, ... = 1/2187?7th term
Q5a₈ = 192, r = 2, find a₁₂a₁₂ = 3072
Q6a₄ = 2/3, a₇ = 16/81a = 9/4, r = 2/3
Q7Two GPs with equal nth termn = 5

What This Exercise Prepares You For

Exercise 6.5 builds the essential skills needed for the rest of Chapter 6, particularly the upcoming exercises on the sum of n terms of a GP and applications of progressions to real-life problems like compound interest and population growth. A strong grip on the nth term formula here makes those sum formulas much easier to understand and apply.

This exercise also connects naturally to Introduction to Progressions, where the foundational concepts of sequences, arithmetic progressions, and geometric progressions are introduced. For Class 9 students revisiting exponent rules used heavily in this exercise, refer back to Exponents and Powers, since comparing powers with the same base is the central technique in Questions 4 and 7.

📐 Board Exam Strategy (Telangana & AP SSC, CBSE): Questions like Q5, Q6, and Q7 — involving two given terms or two GPs — are common 4-mark and 5-mark questions in board exams. Always set up your equations clearly labeled (1) and (2), divide to eliminate 'a', solve for r first, then substitute back. This stepwise method earns full marks even if the final numeric answer has a small error.
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