On this page
In Class 9 you learn to turn 0.666... into 2/3 by multiplying by 10 and subtracting. It works, but it can feel like a trick. In Class 11 and in JEE, the same question comes back as a geometric progression, and then the trick has a reason. This post shows how to write any recurring decimal as an infinite GP, sum it, and avoid the slips that cost marks.
Where does this topic sit in your books and exams?
The new Class 9 textbook, Ganita Manjari, Chapter 3 "The World of Numbers", has a section on terminating and repeating decimals. It converts pure repeating decimals (0.454545... = 45/99 = 5/11) and mixed ones such as 0.1666... = 1/6 by multiplying by powers of 10 and subtracting. It also asks you to explain why 0.999... is exactly 1.
The NCERT Class 11 chapter on Sequences and Series (2026-27 reprint) names "infinite geometric series" and derives the sum of the first n terms, Sn = a(1 − rⁿ)/(1 − r). The JEE (Advanced) 2026 information brochure goes one step further and lists "infinite geometric series" by name in its algebra syllabus. The JEE (Main) 2026 syllabus lists arithmetic and geometric progressions. Check your own year's syllabus, since both are reissued each cycle.
Why does an infinite GP have a finite sum?
Start from the formula NCERT gives: Sn = a(1 − rⁿ)/(1 − r). If −1 < r < 1, then rⁿ gets closer and closer to 0 as n grows. For r = 1/100, r³ is already 0.000001. So the sum of all the terms settles at
S = a/(1 − r), valid only when |r| < 1.
A recurring decimal always has |r| < 1, because each repeat of the block is shifted further to the right by a power of 10.
Worked example 1: a pure recurring decimal
Write 0.363636... as a fraction.
Split it into blocks: 0.36 + 0.0036 + 0.000036 + ...
As fractions: 36/100 + 36/10000 + 36/1000000 + ...
So a = 36/100 and r = 1/100. The block has two digits, so r has two zeros.
S = (36/100) ÷ (1 − 1/100) = (36/100) × (100/99) = 36/99 = 4/11.
Check: 4 ÷ 11 = 0.3636..., as required. The pattern is general: a repeating block of k digits gives r = 1/10ᵏ, and the answer is the block over k nines.
Worked example 2: a decimal with a non-repeating part
Write 1.24545... (45 repeating, the 2 does not) as a fraction.
Separate the parts: 1.2 + 0.04545...
The recurring part is 0.045 + 0.00045 + ..., so a = 45/1000 and r = 1/100.
Its sum is (45/1000) ÷ (99/100) = 45/990 = 1/22.
Total = 12/10 + 1/22 = 132/110 + 5/110 = 137/110.
Check: 137 ÷ 110 = 1.24545... Ganita Manjari reaches answers like this by multiplying by 10 and then by 100 and subtracting. The GP method gives the same fraction and tells you why the subtraction removes the endless tail.
Worked example 3: why 0.999... equals 1
0.999... = 9/10 + 9/100 + 9/1000 + ..., with a = 9/10 and r = 1/10. The sum is (9/10) ÷ (9/10) = 1. There is no gap between 0.999... and 1, because the gap after n nines is 1/10ⁿ, and that shrinks to zero. Ganita Manjari makes the same point: any terminating decimal has a second form ending in repeating 9s, such as 2.47 = 2.46999....
What does a JEE-style question look like?
Exam questions rarely ask for one conversion. They combine several.
Example. Find 0.2323... + 0.3232... as a fraction.
Each is a two-digit block, so each is the block over 99: 23/99 + 32/99 = 55/99 = 5/9. Check: 5/9 = 0.5555..., and 0.2323... + 0.3232... = 0.5555..., digit by digit.
A second type asks for the condition on r. If a question gives a GP with r = 3/2 and asks for "the sum to infinity", the right answer is that it does not exist, because |r| > 1 and the terms grow. If you are unsure about GP basics such as the nth term, our guide to sequences and series for JEE covers them, and arithmetico-geometric series uses the same a/(1 − r) idea.
Which mistakes cost marks?
Wrong ratio for a long block. For 0.142857142857..., the block has six digits, so r = 1/1000000, not 1/10.
Forgetting the non-repeating digits. In 1.24545..., the 2 is not part of the GP. Peel it off first.
Using a/(1 − r) when |r| ≥ 1. The formula has a condition. Write it every time.
Not simplifying. 36/99 is correct but 4/11 is the expected final form. Divide by the HCF.
Skipping the check. A quick long division of your fraction catches almost every error.
A practice routine
Spend 20 minutes on this, three times this week.
Convert five pure recurring decimals, with block lengths 1, 2, 3 and 6, using the GP method. Write a and r before anything else.
Convert five mixed decimals. Underline the non-repeating part first.
Redo two of them with the Class 9 multiply-and-subtract method and confirm you get the same fraction.
Write one line explaining why 0.999... = 1, without looking at this post.
Finish with three sum-to-infinity questions where you must first decide whether the sum exists.
When you get one wrong and cannot see why, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can find the exact line where your a or r went wrong.
Tonight, convert 0.123123... and 2.0909... into fractions, and check both by long division.
