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.
The sum of the sequence 7,77,777,… to n terms is found by rewriting each term as 97(10k−1) and summing two separate series. The final result is 817(10n+1−9n−10).
This problem is a classic example of a geometric progression in disguise. At first glance, the terms 7,77,777,… don't look like a standard GP — the ratio between successive terms isn't constant. But the pattern is clear: each term is a string of 7's. The trick is to connect each term to powers of 10.
Think about it: 7=7×1, 77=7×11, 777=7×111, and so on. The numbers 1,11,111,… themselves aren't a GP either. But notice: 1=910−1, 11=9102−1, 111=9103−1. That's the key — each term becomes a simple expression involving 10k, and then we can sum the resulting geometric series.
Let's work it out step by step.
Rewrite the general term.
The k-th term of the sequence is a number with k digits, all 7's.
Tk=777…7(k digits)=7×111…1(k ones)
And 111…1 (k ones) = 910k−1. So:
Tk=7⋅910k−1=97(10k−1)
Write the sum to n terms.
Let Sn=T1+T2+⋯+Tn. Then:
Sn=97∑k=1n(10k−1)=97(∑k=1n10k−∑k=1n1)
Sum the geometric series.∑k=1n10k is a GP with first term 10, common ratio 10, and n terms. Its sum is:
Q.Fill in the blank: If −72,x,−27 are consecutive terms of a geometric progression, then the value of x will be ______.
›Reveal solutionSolution
If a,x,c are consecutive terms of a G.P., then x2=a⋅c; here that gives x=±1, and x=−1 is the value consistent with a single common ratio across the progression.
For three consecutive G.P. terms a,x,c: ax=xc⇒x2=ac.
Here a=−72, c=−27, so x2=(−72)(−27)=1, giving x=±1.
Checking x=−1: common ratio r=ax=−2/7−1=27, and indeed x⋅r=−1×27=−27=c✓ — a single positive ratio 27 carries all three (negative) terms consistently.