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Some series look like a GP until you notice the numbers in front: 1 + 2x + 3x² + 4x³ + …, or 1 + 4/5 + 7/25 + 10/125 + …. Each term is an AP term times a GP term, so neither the AP nor the GP formula applies. The good news is that the step you already know from the GP proof, multiply by r and subtract, turns these into a plain GP. This post shows that step on finite and infinite series, with checked numbers and the slips that cost marks.
Is this in the syllabus?
The JEE (Advanced) 2026 brochure lists "sums of finite arithmetic and geometric progressions, infinite geometric series". The JEE (Main) 2026 syllabus lists "Arithmetic and Geometric progressions". Neither names arithmetico-geometric series. NCERT's Class 11 chapter on sequences and series derives the GP sum by multiplying by r and subtracting, which is the whole method here. Treat this as an application of the GP sum, and check your exam's current syllabus. Our sequences and series guide covers the AP and GP sums themselves.
What is an arithmetico-geometric series?
Its k-th term is [a + (k − 1)d] × rᵏ⁻¹: the first factor runs along an AP (a, a + d, a + 2d, …) and the second along a GP (1, r, r², …). For 1 + 4/5 + 7/25 + 10/125 + …, the AP part is 1, 4, 7, 10 (a = 1, d = 3) and the GP part has r = 1/5.
How do you sum a finite one?
Example 1. Find S = 1·2 + 2·2² + 3·2³ + … + n·2ⁿ.
Multiply by r = 2 and shift each term one place right:
S = 2 + 2·4 + 3·8 + … + n·2ⁿ
2S = 4 + 2·8 + … + (n − 1)·2ⁿ + n·2ⁿ⁺¹
Subtract the second line from the first. Matching terms leave 2, 4, 8, …, 2ⁿ, a GP, and the last term of 2S is left over:
S − 2S = (2 + 4 + … + 2ⁿ) − n·2ⁿ⁺¹ = (2ⁿ⁺¹ − 2) − n·2ⁿ⁺¹.
So −S = (1 − n)2ⁿ⁺¹ − 2, which gives S = (n − 1)2ⁿ⁺¹ + 2.
Check with n = 4: 2 + 8 + 24 + 64 = 98, and (3)(32) + 2 = 98.
How do you sum an infinite one?
For |r| < 1, the leftover term shrinks to zero, so the same steps give a clean answer.
Example 2. Find S = 1 + 4/5 + 7/25 + 10/125 + ….
S = 1 + 4/5 + 7/25 + 10/125 + …
S/5 = 1/5 + 4/25 + 7/125 + …
Subtracting: (4/5)S = 1 + 3/5 + 3/25 + 3/125 + … = 1 + 3 × (1/5)/(1 − 1/5) = 1 + 3/4 = 7/4. So S = 35/16.
The general result is S = a/(1 − r) + dr/(1 − r)². With a = 1, d = 3 and r = 1/5 it gives 5/4 + (3/5)/(16/25) = 5/4 + 15/16 = 35/16, which agrees. Derive it once; do not rely on remembering it.
Example 3. 1/3 + 2/9 + 3/27 + … has terms k/3ᵏ. Subtracting one-third of the sum leaves (2/3)S = 1/3 + 1/9 + 1/27 + … = 1/2, so S = 3/4.
Where else does the trick show up?
Power series. 1 + 2x + 3x² + … = 1/(1 − x)² for |x| < 1. At x = 1/2 the sum is 4.
Probability. If you toss a fair coin until the first head, the expected number of tosses is 1·(1/2) + 2·(1/4) + 3·(1/8) + …. This is half of the series above at x = 1/2, so the answer is 2. The JEE (Advanced) 2026 brochure lists the mean of a random variable.
Which mistakes cost the most marks?
Dropping the leftover term in a finite sum, such as −n·2ⁿ⁺¹ in example 1.
Misaligning the lines. Shift rS one place right before subtracting, so that like powers sit together.
Forgetting the first term stands outside the new GP, like the 1 in example 2.
Using the infinite result when |r| ≥ 1. The series 1 + 2·2 + 3·4 + … has no finite sum.
Sign slips with negative r. For 1 − 2x + 3x² − …, the sum is 1/(1 + x)².
A practice routine
Re-deriving from memory beats rereading. In Roediger and Karpicke (2006), students who took recall tests on passages retained more on delayed tests than students who restudied, even though restudying made students more confident.
Derive, do not copy. On a blank page, sum 1 + 2r + 3r² + … from scratch each day for a week.
Check every finite answer by adding the first four terms by hand.
Vary one thing at a time: change d, then r, then make r negative.
Pair it with the method of differences, the other tool for series that are not AP or GP, in our post on summing series term by term.
If you get stuck on the alignment step, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, sum 1 + 3x + 5x² + 7x³ + … for |x| < 1 on a blank page, then check it at x = 1/2.
