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Common Mistakes

Traps in Sequence & Series

5 mistake patterns students fall for. 2 high-frequency traps appear in almost every exam.

Off-by-one in AP term numbering

Very CommonFORMULA

Using an=a+nda_n = a + nd instead of a+(n−1)da + (n-1)d. The 1st term is aa, not a+da + d.

Why: The index starts at 1, not 0, so the first term has zero common differences added.

WRONG: 5th term =a+5d= a + 5d
RIGHT: 5th term =a+4d= a + 4d (always subtract 1 from the term number)
See pattern: AP Sum & Term Finding →

Forgetting r=1r = 1 case in GP

Very CommonDOMAIN

The GP sum formula Sn=a(rn−1)r−1S_n = \frac{a(r^n - 1)}{r - 1} is undefined when r=1r = 1.

Why: The formula has (r−1)(r - 1) in the denominator, which is zero when r=1r = 1.

WRONG: Applying the formula blindly and getting 00\frac{0}{0}
RIGHT: When r=1r = 1, all terms equal aa, so Sn=naS_n = na. Always check r=1r = 1 separately.
See pattern: GP Properties & Sum →

Infinite GP applied when ∣r∣≥1|r| \geq 1

CommonDOMAIN

S∞=a1−rS_\infty = \frac{a}{1 - r} only works when ∣r∣<1|r| < 1. Series diverges otherwise.

Why: Students memorize the formula without the convergence condition.

WRONG: Sum of 1+2+4+8+⋯=11−2=−11 + 2 + 4 + 8 + \cdots = \frac{1}{1 - 2} = -1
RIGHT: This GP diverges since ∣r∣=2>1|r| = 2 > 1. No finite sum exists.
See pattern: Infinite GP & Convergence →

Confusing SnS_n with ana_n

CommonFORMULA

Students sometimes use the SnS_n formula when asked for ana_n, or vice versa.

Why: Both involve the same variables aa, dd, and nn, so the formulas look similar.

WRONG: Asked for the 10th term, computing S10S_{10} instead
RIGHT: an=Sn−Sn−1a_n = S_n - S_{n-1}. The nnth term is the difference of consecutive partial sums.
See pattern: AP Sum & Term Finding →

Sign error in method of differences

CommonSIGN ERROR

When computing Tn=Sn−Sn−1T_n = S_n - S_{n-1}, students mess up the algebra while expanding.

Why: Expanding (n−1)2(n-1)^2 requires careful distribution of the negative sign across all terms.

WRONG: Sn=n2+nS_n = n^2 + n, so Tn=n2+n−(n−1)2−(n−1)=2nT_n = n^2 + n - (n-1)^2 - (n-1) = 2n (wrong expansion)
RIGHT: Tn=(n2+n)−((n−1)2+(n−1))=n2+n−n2+2n−1−n+1=2nT_n = (n^2 + n) - ((n-1)^2 + (n-1)) = n^2 + n - n^2 + 2n - 1 - n + 1 = 2n
See pattern: Special Series (Method of Differences) →
Test yourself

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