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 fifth-term condition fixes the common ratio, then the sum condition fixes the first term for each ratio. The two possible G.P.s are −34,−38,−316,… (ratio 2) and 4,−8,16,… (ratio −2).
A G.P. with first term a and common ratio r has terms a,ar,ar2,ar3,ar4,…
The problem gives two conditions:
Sum of the first two terms: a+ar=−4
Fifth term is 4 times the third term: ar4=4(ar2)
The second condition involves only r (the a cancels out), so it makes sense to solve for r first.
Step 1, find r:
ar4=4ar2⟹r2=4⟹r=2orr=−2(a=0,r=0)
Watch out
Dividing by ar2 is valid here because a=0 (otherwise the sum condition a(1+r)=−4 could never hold) and r=0 (otherwise the third and fifth terms would both be 0, which does not satisfy this problem).
Step 2, find a for each value of r, using a(1+r)=−4:
If r=2: a(1+2)=3a=−4, so a=−34
If r=−2: a(1−2)=−a=−4, so a=4
Step 3, write out and check each G.P.:
a=−34, r=2: the G.P. is −34,−38,−316,−332,−364,… …