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8.3 Geometric Progression (G.P.)

🎓 Class 11 Mathematics CBSE Theory Ch 8 — Sequences and Series ⏱ ~15 min
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This MCQ module is based on: 8.3 Geometric Progression (G.P.)

This mathematics assessment will be based on: 8.3 Geometric Progression (G.P.)
Targeting Class 11 level in Sequences Series, with Advanced difficulty.

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8.3 Geometric Progression (G.P.)

You met arithmetic progressions in Class 9 — sequences where each term differs from the previous by a fixed amount \(d\) (the common difference): \(a, a+d, a+2d, \ldots\). The natural multiplicative analogue is a geometric progression?.

Geometric Progression (G.P.)
A sequence \(a_1, a_2, a_3, \ldots\) is a geometric progression if every term (after the first) bears the same ratio to the previous term. The fixed ratio is the common ratio \(r\): \[\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=\dfrac{a_4}{a_3}=\cdots=r.\] Equivalently, \(a_n=a_{n-1}\cdot r\) for \(n\ge 2\), with \(a_1=a\) (the first term).

Examples of G.P.:

  • \(2, 4, 8, 16, 32, \ldots\) with \(a=2\), \(r=2\).
  • \(\tfrac{1}{9}, -\tfrac{1}{27}, \tfrac{1}{81}, \ldots\) with \(a=1/9\), \(r=-1/3\).
  • \(0.01, 0.0001, 0.000001, \ldots\) with \(a=0.01\), \(r=0.01\).

Counter-examples (NOT G.P.): \(2, 4, 7, 11, \ldots\) — ratios \(4/2=2,\ 7/4=1.75\), not constant.

Formula for the nth term

Starting from \(a_1=a\):

\[a_2=a\cdot r,\quad a_3=a\cdot r^2,\quad a_4=a\cdot r^3,\quad \ldots\]

By induction:

nth term of a G.P.
\[\boxed{\;a_n=a\,r^{\,n-1}\;}\] where \(a\) is the first term and \(r\) is the common ratio. Setting \(n=1\) recovers \(a_1=a\); the exponent \(n-1\) is one less than the index because the first term is already given.

Three useful identities

Properties of a G.P.
  1. Geometric-mean characterisation: \(a_n^2=a_{n-1}\cdot a_{n+1}\) — every interior term is the geometric mean of its neighbours.
  2. Equidistant terms: \(a_{n-k}\cdot a_{n+k}=a_n^2\) for any \(k\) — pairs of terms equidistant from a central term have constant product.
  3. Multiplying every term by a constant: if \((a_n)\) is a G.P. with ratio \(r\) and \(c\ne 0\), then \((c\,a_n)\) is also a G.P. with the same ratio \(r\).
n123456
aₙ for a=2, r=3261854162486
aₙ for a=1, r=1/211/21/41/81/161/32
aₙ for a=2, r=−32−618−54162−486

Worked Examples

Example 5. Find the 10th and \(n\)th terms of the G.P. 5, 25, 125, … .
\(a=5\), \(r=25/5=5\). \(a_n=5\cdot 5^{n-1}=5^n\). So \(a_{10}=5^{10}=9\,765\,625\).
Example 6. Which term of the G.P. 2, 8, 32, … equals 131072?
\(a=2,\ r=4\). \(a_n=2\cdot 4^{n-1}=131072\Rightarrow 4^{n-1}=65536=4^8\Rightarrow n-1=8\Rightarrow n=9\). So \(a_9=131072\).
Example 7. Find the G.P. whose 3rd term is 24 and 6th term is 192.
\(a_3=ar^2=24\), \(a_6=ar^5=192\). Divide: \(r^3=192/24=8\Rightarrow r=2\). Then \(ar^2=4a=24\Rightarrow a=6\). G.P.: 6, 12, 24, 48, 96, 192, … .
Example 8. The 5th, 8th and 11th terms of a G.P. are p, q, s respectively. Show that \(q^2=ps\).
\(p=ar^4,\ q=ar^7,\ s=ar^{10}\). Then \(q^2=a^2r^{14}\) and \(ps=a^2r^{14}\). Hence \(q^2=ps\). \(\square\) (This is the equidistant-terms identity \(a_n^2=a_{n-k}a_{n+k}\) with \(n=8, k=3\).)
Activity: Spot the Common Ratio
L3 Apply
Materials: Calculator.
Predict: Is the sequence 1, 2, 6, 24, 120, 720 a G.P.? If yes, find r; if no, what kind of pattern is it?
  1. Compute ratios: 2/1 = 2, 6/2 = 3, 24/6 = 4, 120/24 = 5, 720/120 = 6.
  2. Ratios are 2, 3, 4, 5, 6 — NOT constant. So it's NOT a G.P.
  3. This is the factorial sequence: \(a_n=n!\). The "ratio of consecutive terms" itself grows arithmetically.
  4. Now try 3, 12, 48, 192, 768. Ratios: 4, 4, 4, 4. ✓ G.P. with \(a=3,\ r=4\).
  5. Try -1, 2, -4, 8, -16. Ratios: -2, -2, -2, -2. ✓ G.P. with \(a=-1,\ r=-2\).
A sequence is a G.P. iff every consecutive ratio is the same constant r. The factorial sequence n! is NOT a G.P. — its ratios grow. The Fibonacci sequence is also NOT a G.P., but its ratios converge to the golden ratio φ ≈ 1.618. So Fibonacci is "asymptotically geometric" with ratio φ — a useful approximation for large n.

Competency-Based Questions

Scenario: A car loses value at 15% per year due to depreciation. Bought for ₹8,00,000 in year 0.
Q1. The value of the car at the end of year n is given by:
L3 Apply
  • (a) 800000(1 + 0.15)ⁿ
  • (b) 800000 − 0.15n
  • (c) 800000(0.85)ⁿ
  • (d) 800000/(0.15n)
Answer: (c). Each year the value is multiplied by 0.85 (loses 15%). G.P. with a = 800000, r = 0.85.
Q2. After how many full years will the value first drop below ₹4,00,000?
L4 Analyse
Answer: Need \(800000\cdot 0.85^n<400000\), i.e. \(0.85^n<0.5\). Take log: \(n\log 0.85<\log 0.5\), \(n>\log 0.5/\log 0.85\approx 4.27\). So at year 5 (after 5 years), value first dips below ₹4 lakh.
Q3. (Fill in) The 10th term of the G.P. 4, 8, 16, … is ____ .
L3 Apply
Answer: a = 4, r = 2. \(a_{10}=4\cdot 2^9=2048\).
Q4. (T/F) "Every G.P. with positive first term and r > 1 grows unboundedly." Justify.
L5 Evaluate
True. \(a_n=ar^{n-1}\to\infty\) as \(n\to\infty\) because \(r^{n-1}\to\infty\) when \(r>1\). For \(0
Q5. Design: insert two numbers between 3 and 81 so that the resulting four-term sequence is a G.P. Find the inserted numbers.
L6 Create
Solution: 3, x, y, 81 in G.P. gives 81 = 3·r³, so r³ = 27, r = 3. Then x = 3·3 = 9, y = 9·3 = 27. So sequence is 3, 9, 27, 81. (Verify ratios are all 3 ✓.)

Assertion–Reason Questions

Assertion (A): The sequence 1, 1, 1, 1, … is a G.P.
Reason (R): Each consecutive ratio equals 1.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). Constant common ratio (1) makes it a G.P. (also an A.P. with d = 0 — degenerate but valid in both senses).
Assertion (A): If \(a, b, c\) are in G.P., then \(b^2=ac\).
Reason (R): The middle term of a G.P. is the geometric mean of its neighbours.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). R is the named property; A is its formal statement.
Assertion (A): Common ratio of 4, −2, 1, −1/2, … is −1/2.
Reason (R): r = a₂/a₁ = −2/4 = −1/2.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). Verified by computing the first ratio.

Frequently Asked Questions

What is a geometric progression?
A sequence a₁, a₂, a₃, … is a geometric progression (G.P.) if the ratio of any term to its predecessor is the same constant r (the common ratio). Equivalently, aₙ = aₙ₋₁ · r for n ≥ 2.
What is the formula for the nth term of a G.P.?
If the first term is a and common ratio is r, then aₙ = a · rⁿ⁻¹.
What is the common ratio of a geometric progression?
The fixed number r = aₙ₊₁ / aₙ that you multiply each term by to get the next term.
Can the common ratio be negative?
Yes. If r < 0, the terms alternate in sign. For example, 2, −6, 18, −54, … has a = 2 and r = −3.
How is a G.P. different from an A.P.?
In an A.P. successive terms differ by the same constant d (added). In a G.P. successive terms have the same ratio r (multiplied). A.P. grows linearly; G.P. grows exponentially.
What is the relationship between consecutive terms in a G.P.?
Each term is the geometric mean of its neighbours: aₙ² = aₙ₋₁ · aₙ₊₁.
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