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,ar2,ar3,ar4,…
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:
Tn=a⋅rn−1
where Tn 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⋅25−1=24=16, which matches our list.
Sum of n Terms
There are two cases, depending on whether r=1 or not.
Sum of first n terms of a GP:
Sn=⎩⎨⎧a⋅r−1rn−1,n⋅a,r=1r=1
When r=1, every term is just a, so the sum is simply n×a.
Why the formula works (intuition):
Let S=a+ar+ar2+⋯+arn−1. Multiply both sides by r: rS=ar+ar2+⋯+arn. Subtract the first from the second: rS−S=arn−a, so S(r−1)=a(rn−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∞=1−ra,for ∣r∣<1
Watch out
If ∣r∣≥1, the infinite sum does not exist (it diverges to infinity or oscillates without settling). Never apply the infinite sum formula when ∣r∣≥1.
Example:1+21+41+81+… has a=1, r=21, so S∞=1−1/21=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=TnTn+1
Always
n-th term
Tn=arn−1
Always
Sum of n terms
Sn=ar−1rn−1
r=1
Sum of n terms
Sn=na
r=1
Infinite sum
S∞=1−ra
$
Common Mistakes to Avoid
Confusing n and n−1: The first term corresponds to n=1, so the exponent is n−1, not n.
Using infinite sum when ∣r∣≥1: The formula gives a finite number, but the actual sum is infinite — it's a trap.
Forgetting the sign when r is negative: Terms alternate, and the sum formula still works, but be careful with signs in calculations.
Why This Matters
Geometric progressions appear everywhere: compound interest in finance, population growth in biology, radioactive decay in physics, and even in the design of algorithms (binary search halves the problem size each step — a GP with r=1/2). Once you see the pattern of repeated multiplication, you'll spot GPs in many real-world contexts.
Geometric Progression is one of the two central sequence types in the NCERT Class 11 Mathematics chapter on Sequences and Series, and searches like "geometric progression: definition, formula and examples" or "GP sum of n terms important questions" point straight to this concept. It's also a regular fixture in JEE Main, CET, and other competitive exams, especially problems involving compound interest and infinite series.
Concept: Geometric Progression – each term is obtained by multiplying the previous term by a fixed common ratio r.
Here, first term a=2, common ratio r=28=4.
The k-th term of a G.P. is ark−1.
We set 2⋅4k−1=131072.
Divide both sides by 2: 4k−1=65536.
Write 65536 as a power of 4: 48=65536 (since 48=(22)8=216=65536).
Thus k−1=8, so k=9.
✓Final answer
The 9th term of the G.P. is 131072.
In a geometric progression, each term is the first term multiplied by the common ratio raised to the power (n−1). For the G.P. 2,8,32,…, the common ratio is 4, and solving 2⋅4k−1=131072 gives k=9. So the 9th term is 131072.
A geometric progression (G.P.) is a sequence where each term after the first is obtained by multiplying the previous term by a fixed number called the common ratio (r). The key idea is that the terms grow (or shrink) by a constant factor, not a constant difference.
Here, the sequence is 2,8,32,….
To find the common ratio, divide any term by the one before it:
8÷2=4, and 32÷8=4. So r=4.
The general term (the k-th term) of a G.P. with first term a and common ratio r is given by:
Tk=a⋅rk−1
For this problem, a=2, r=4, and we want Tk=131072.
Set up the equation
Substitute the known values into the formula:
2⋅4k−1=131072
Isolate the power
Divide both sides by 2:
4k−1=2131072=65536
Express both sides as powers of the same base
Since 4=22, we can rewrite 4k−1 as (22)k−1=22(k−1).
Now, 65536 is a power of 2. Let's find which one.
You can check by repeated doubling: 210=1024, 211=2048, 212=4096, 213=8192, 214=16384, 215=32768, 216=65536.
So 65536=216.
Tip
A faster way: 65536=64×1024=26×210=216. Knowing powers of 2 up to 210 is very handy for such problems.
Equate the exponents
Now we have:
22(k−1)=216
Since the bases are equal (and 2=1), the exponents must be equal:
2(k−1)=16
Solve for k
Divide both sides by 2:
k−1=8
So k=9.
Watch out
A common mistake is to write Tk=ark instead of ark−1. The first term corresponds to k=1, so the exponent must be k−1 to give r0=1. If you used k instead, you'd get k=10, which is off by one.
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 2k (2 parents, 4 grandparents, 8 great-grandparents, ...). Over 10 generations, the total is the geometric series:
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2026Set ANNUAL1 mark
Q.If x+9,x−6,4 are in G.P, then what are the values of x?
›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: x2−12x+36=4x+36
x2−12x−4x=0
x2−16x=0
x(x−16)=0
So x=0 or x=16.
Check x=0: terms are 9, −6, 4, ratio −6/9=−2/3 and 4/(−6)=−2/3 — consistent, valid G.P.
Check x=16: terms are 25, 10, 4, ratio 10/25=2/5 and 4/10=2/5 — consistent, valid G.P.
Both values work.
✓Final answer
x = 0 or x = 16.
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2025Set ANNUAL1 markMCQ
Q.The sum of infinity of the G.P. 1,31,91,… is
(a) Infinity
(b) 0
(c) 31
(d) 23
›Reveal solutionSolution
For the G.P. 1,31,91,… with a=1, r=31, the infinite sum is S∞=1−ra=23.
Here a=1 and the common ratio r=11/3=31. Since ∣r∣=31<1, the series converges and
S∞=1−ra=1−311=321=23.
✓Final answer
The sum to infinity is 23 — option (d).
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2025Set ANNUAL1 markMCQ
Q.Questions numbers 11 and 12 are assertion and reason based questions. Two statements are given, one lebelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below.
Assertion (A): For x=±1, the numbers 7−2,x,2−7 are in G.P.
Reason (R): Three numbers a, b, c are in G.P. if b2=ac.
(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 b2=ac to −72,x,−27 gives x2=1, i.e. x=±1, 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: b2=ac — this is exactly Reason (R), which is a true, standard fact about G.P.s.
Apply it to −72,x,−27: treating x as the middle term,
x2=(−72)(−27)=1⟹x=±1.
This confirms Assertion (A) is true, and it is true precisely BECAUSE of the G.P. rule stated in (R) — so (R) is the correct explanation of (A).
✓Final answer
Option (a): both Assertion (A) and Reason (R) are true, and (R) is the correct explanation of (A).
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2023Set ANNUAL1 markMCQ
Q.If an+bnan+1+bn+1 is the G.M. between a and b, then the value of n is
(a) 21
(b) −21
(c) 1
(d) -1
›Reveal solutionSolution
The value of n that makes (an+1+bn+1)/(an+bn) equal to the geometric mean ab of a and b is n=−21.
The G.M. between a and b is ab. We are told
an+bnan+1+bn+1=ab
Step 1: Try n=−21.
a−1/2+b−1/2a1/2+b1/2
Multiply numerator and denominator by ab:
aba−1/2+abb−1/2ab(a1/2+b1/2)=b+aab(a+b)
The factor (a+b) cancels top and bottom, leaving exactly ab — confirming n=−21 works.
Step 2: Quick numeric check.
Take a=1,b=4, so ab=2. With n=−21:
1−1/2+4−1/211/2+41/2=1+0.51+2=1.53=2
This matches ab=2 exactly.
✓Final answer
The correct option is (b) −21.
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2022Set ANNUAL1 mark
Q.For what values of x are the numbers −72,x,−27 in G.P.?
›Reveal solutionSolution
Using x2=(−72)(−27) gives x2=1, so x=±1.
Three numbers a,b,c are in G.P. exactly when b2=ac (the middle term is the geometric mean of the outer two).