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Questions like "find the sum of 3 + 33 + 333 + … to n terms" look as if they need a new formula. They do not. Every series of repeated digits hides a geometric progression, and one rewriting step brings it out. This post shows that step, works four series with checks, and lists the traps.
Where is this in the syllabus?
NCERT's Class 11 chapter on sequences and series (2026-27 reprint) has this exact idea in Example 10, which sums 7 + 77 + 777 + … by writing it as 7/9 × (9 + 99 + 999 + …). Its miscellaneous exercise then asks for 5 + 55 + 555 + … and 0.6 + 0.66 + 0.666 + …. The JEE (Main) 2026 syllabus lists "Arithmetic and Geometric progressions" in Unit 6, and the CBSE 2026-27 curriculum lists the sum of n terms of a GP in Class 11. To keep your practice fresh, the examples below avoid the NCERT numbers. Check your own year's syllabus.
Why is 9 + 99 + 999 not a GP?
A GP has a constant ratio. Here 99/9 = 11, but 999/99 ≈ 10.09. The ratio changes, so the GP formula cannot be used directly. What does stay simple is each term:
9 = 10 − 1
99 = 10² − 1
999 = 10³ − 1
the k-th term, k nines, is 10ᵏ − 1.
So the sum of n terms is (10 + 10² + … + 10ⁿ) − (1 + 1 + … + 1). The first bracket is a GP with a = 10 and r = 10. The second is just n.
Worked example 1: 9 + 99 + 999 + … to n terms
GP part: 10(10ⁿ − 1)/(10 − 1) = 10(10ⁿ − 1)/9.
Subtract n: Sₙ = 10(10ⁿ − 1)/9 − n, or (10ⁿ⁺¹ − 10)/9 − n.
Check with n = 3: 10 × 999/9 − 3 = 1110 − 3 = 1107. Directly, 9 + 99 + 999 = 1107.
Worked example 2: 3 + 33 + 333 + … to n terms
Each term is a third of the matching term above: 33 = 99/3. So:
Sₙ = (1/3)[10(10ⁿ − 1)/9 − n].
Over a common denominator: Sₙ = (10ⁿ⁺¹ − 10 − 9n)/27.
Check with n = 4: (100,000 − 10 − 36)/27 = 99,954/27 = 3702. Directly, 3 + 33 + 333 + 3333 = 3702.
Any digit d works the same way: d + dd + ddd + … = (d/9)[10(10ⁿ − 1)/9 − n].
Worked example 3: 1 + 11 + 111 + … to n terms
With d = 1: Sₙ = (10ⁿ⁺¹ − 10 − 9n)/81. For n = 5: (1,000,000 − 10 − 45)/81 = 999,945/81 = 12,345. The neat digits are a good sign your formula is right: 1 + 11 + 111 + 1111 + 11111 does equal 12,345.
Worked example 4: 0.4 + 0.44 + 0.444 + … to n terms
Decimals need one change. 0.9 = 1 − 1/10, 0.99 = 1 − 1/10², and so on, so 0.444… with k fours is (4/9)(1 − 10⁻ᵏ).
Sₙ = (4/9)[n − (1/10 + 1/10² + … + 1/10ⁿ)].
The bracket is a GP with a = 1/10 and r = 1/10, whose sum is (1/9)(1 − 10⁻ⁿ).
Sₙ = (4/9)[n − (1/9)(1 − 10⁻ⁿ)].
Check with n = 3: (4/9)(3 − 0.111) = (4/9)(2.889) = 1.284. Directly, 0.4 + 0.44 + 0.444 = 1.284.
Here r is 1/10, not 10. Students who carry r = 10 over from the whole-number case get a sum that grows with every power of 10, which cannot be right for terms all below 1.
If you want the infinite version of the decimal GP, our post on recurring decimals as infinite GPs covers it. For the GP formula itself, see sequences and series for JEE.
Which mistakes cost the most marks?
Treating the series as a GP. Test the ratio of two pairs of terms first.
Forgetting the −n. Every term contributes a −1, and there are n of them.
Wrong first term in the GP part. It is 10, not 1, so the GP sum is 10(10ⁿ − 1)/9.
Using r = 10 for decimals. For 0.4 + 0.44 + …, the ratio is 1/10.
Skipping the check. Put n = 1 and n = 2 into your answer. For example 2, n = 1 gives (100 − 10 − 9)/27 = 3. Correct.
A practice routine
Dunlosky and colleagues (2013) rated practice testing and distributed practice as the two high-utility study techniques in their review. Use both: short tests, spread out.
Session 1 (15 minutes): sum 2 + 22 + 222 + … and 8 + 88 + 888 + … to n terms. Check each with n = 3.
Session 2, two days later: 0.7 + 0.77 + 0.777 + … and 0.5 + 0.55 + … to n terms.
Session 3, a week later: from blank paper, write the general rule for d + dd + ddd + … and test it on one example from each session.
If a series will not split and you cannot see why, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can compare its first step with yours.
Tonight, find 6 + 66 + 666 + … to n terms and check your formula with n = 2.
