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CAT (Common Admission Test) · Advanced Quantitative Aptitude

Sequences and Series

Arithmetic, geometric and harmonic progressions and summation of series.

Seven concepts. Arithmetic, geometric and harmonic progressions, the summation formulas CAT expects you to recall, and telescoping — each with a ledger traced from the legacy examples.

  • CAT (Common Admission Test)
  • Medium level
  • 7 concepts
  • 21 practice questions

1Arithmetic progression and the nth term

An arithmetic progression (AP) is a sequence whose consecutive terms differ by a fixed constant d. The first term is a and the n-th term is t_n = a + (n-1)d — read d from any two consecutive terms, not from the first alone.

Any term is the average of two terms equidistant from it: t_k = \frac{t_{k-r} + t_{k+r}}{2}. When three consecutive terms are symmetric about a middle value, write them as a-d,\ a,\ a+d so the common difference cancels in sums.

Figure. Each step along n raises the term by the same d=4, so the graph is a straight line — constant difference is constant slope.

How it works

  1. Read a and dThe first term is a; subtract any term from the next to get d.
  2. Apply the nth-term formulat_n = a + (n-1)d counts (n-1) steps of size d from the start.
  3. Average equidistant partnersA missing middle term is the mean of the two terms the same distance away — faster than solving for d when both ends are given.

The 20th term of an AP

Find the 20th term of the AP 3, 7, 11, \ldots

  • a=3, d=7-3=4, n=20d=4
  • t_{20} = a + (n-1)d = 3 + 19 \times 43 + 76
  • t_{20}79

Pro tip. If the 5th and 15th terms are given, the 10th is their average — do not re-derive d first unless you need a term that is not halfway between two known ones. The symmetric triple a-d,\ a,\ a+d is the same idea in three consecutive slots.

In an AP the 5th term is 23 and the 15th term is 63. The 10th term is
  1. 43
  2. 46
  3. 40

The 10th term is equidistant from the 5th and 15th, so t_{10} = (23 + 63)/2 = 43. Adding the gap 40/10 = 4 five times from 23 also gives 43, but averaging is one step when the target sits midway.

2Sum of an arithmetic progression

The sum of the first n terms of an AP equals the number of terms times the average of the first and last: S_n = \frac{n}{2}(a + l) where l = t_n.

Expanding that average gives the standard form S_n = \frac{n}{2}\big[2a + (n-1)d\big]. Use whichever version needs fewer inputs — if a and l are already on the page, the half-sum formula is one multiplication.

Figure. For the AP 3,7,11,15,19, every end-pair sums to 22. Five terms make 5/2\times 22=55 — constant pair-sum is why S_n=\frac{n}{2}(a+l) works.

How it works

  1. Fix n, a and either d or lRead n from the wording; compute l = a + (n-1)d if only d is given.
  2. Pick the sum formulaS_n = \frac{n}{2}(a+l) when both ends are known; otherwise \frac{n}{2}[2a+(n-1)d].
  3. Multiply out onceOne bracket expansion — do not re-derive d if l was supplied directly.

Sum of an arithmetic progression

Find the sum of the first 20 terms of the AP 3, 7, 11, \ldots

  • a=3, d=4, n=20d=4
  • S_{20} = \frac{20}{2}[2a + (n-1)d] = 10[6 + 19 \times 4]10[6 + 76]
  • S_{20}820

Pro tip. If both first and last terms are easy, \frac{n}{2}(a+l) is even faster than the full formula — here l = t_{20} = 79, so S_{20} = 10(3+79) = 820 in one line.

The sum of the first 10 terms of the AP 5, 9, 13, \ldots is
  1. 230
  2. 225
  3. 245

a=5, d=4, l = 5 + 9 \times 4 = 41, so S_{10} = \frac{10}{2}(5+41) = 230. Using \frac{10}{2}[10 + 36] = 5 \times 46 also gives 230.

3Geometric progression — nth term and finite sum

A geometric progression (GP) has consecutive terms in a fixed ratio r: the n-th term is t_n = ar^{n-1} where a is the first term.

The sum of the first n terms is S_n = a\frac{r^n - 1}{r - 1} for r \neq 1. When |r| > 1 terms grow in magnitude; when 0 < r < 1 they shrink toward zero — the same formula handles both.

Figure. Each step multiplies by r=3, so heights scale by three — constant ratio is exponential growth on linear axes, not a straight line.

How it works

  1. Read a and rDivide any term by its predecessor; check sign if terms alternate.
  2. Nth term or finite sumt_n = ar^{n-1} for a single term; S_n = a(r^n-1)/(r-1) when adding n terms.
  3. Watch r = 1Equal terms give S_n = na — the GP formula has r-1 in the denominator and does not apply.

The 5th term of a GP

Find the 5th term of the GP 2, 6, 18, \ldots

  • a=2, r=6/2=3, n=5r=3
  • t_5 = ar^{n-1} = 2 \times 3^42 \times 81
  • t_5162

Pro tip. For a finite sum of the same GP, S_5 = 2(3^5-1)/(3-1) = 2 \times 242/2 = 242 — one ratio read, two formulas.

The sum of the first 4 terms of the GP 3, 6, 12, \ldots is
  1. 45
  2. 39
  3. 48

a=3, r=2, so S_4 = 3(2^4-1)/(2-1) = 3 \times 15 = 45. Adding 3+6+12+24 confirms.

4Infinite geometric series

An infinite GP converges to a finite sum only when |r| < 1. Then S_\infty = \frac{a}{1-r}.

If |r| \geq 1 the partial sums grow without bound (or oscillate when r is negative) — there is no finite S_\infty. Always verify the ratio before applying the formula.

Figure. Partial sums climb toward the dashed line S_\infty=16; each new term is half the previous leftover gap to that line.

How it works

  1. Read a and rIdentify the first term and the common ratio from consecutive terms.
  2. Check |r| < 1If the ratio is \frac12, -\frac13, etc., the series converges; if |r| \geq 1, stop — no finite sum.
  3. Apply S_\infty = a/(1-r)One division once convergence is confirmed.

Sum of an infinite GP

Find the sum of the infinite series 8 + 4 + 2 + 1 + \cdots

  • a=8, r=4/8=\frac12; |r|<1 so it convergesr=\frac12
  • S_\infty = \frac{a}{1-r} = \frac{8}{1-\frac12}\frac{8}{1/2}
  • S_\infty16

Pro tip. An infinite GP only has a finite sum when |r|<1; always verify the ratio before applying the formula.

The sum to infinity of 12 + 4 + \frac{4}{3} + \cdots is
  1. 18
  2. 24
  3. 36

a=12, r=\frac13, so S_\infty = 12/(1-\frac13) = 12/(\frac23) = 18. The ratio \frac13 satisfies |r|<1.

5Harmonic progression via reciprocals

A harmonic progression (HP) is a sequence whose reciprocals form an AP. To solve any HP problem, invert every term to get an AP, run AP machinery, then invert back.

If the reciprocals are a, a+d, a+2d, \ldots then the HP terms are \frac1a, \frac1{a+d}, \frac1{a+2d}, \ldots — never apply AP formulas directly to HP terms.

Figure. An HP is an AP after taking reciprocals (scaled in the figure so both series fit). Solve in the AP world, then invert back — never average HP terms directly.

How it works

  1. Take reciprocalsReplace each HP term h_k by 1/h_k to obtain an AP.
  2. Solve the APRead a and d on the reciprocals; use t_n = a+(n-1)d for the target reciprocal.
  3. Invert the answerThe HP term is the reciprocal of the AP term you found.

Fourth term of a harmonic progression

Find the 4th term of the HP whose reciprocals are 3, 6, 9, \ldots

  • Reciprocals form AP: a=3, d=3d=3
  • 4th reciprocal: t_4 = 3 + 3 \times 312
  • 4th HP term = 1/t_4\frac{1}{12}

Pro tip. HP terms themselves never have a constant difference — if subtracting consecutive terms gives messy fractions, you are looking at an HP and should invert first.

The 3rd term of the HP \frac12, \frac14, \ldots is
  1. \frac{1}{6}
  2. \frac{1}{8}
  3. \frac{1}{5}

Reciprocals are 2, 4, 6, \ldots — AP with a=2, d=2. The 3rd reciprocal is 6, so the 3rd HP term is 1/6.

6Standard summation identities

Three closed forms appear often enough to memorise: \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}, \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2, and the link \sum_{k=1}^{n} k^3 = \left(\sum_{k=1}^{n} k\right)^2.

Two shortcuts from the trick sheet: the sum of the first n odd numbers is n^2, and the sum of the first n even numbers is n(n+1). Spot which pattern the question hides before expanding a long sum.

Keep the three closed forms ready: \sum_{k=1}^n k = n(n+1)/2, \sum k^2 = n(n+1)(2n+1)/6, \sum k^3 = [n(n+1)/2]^2. Expand a polynomial sum into these — do not invent a new formula mid-question.

How it works

  1. Classify the sumSquares, cubes, consecutive odds or evens — each has a named closed form.
  2. Plug in nCount how many terms the wording gives; that count is n.
  3. Cross-check small nFor n=3, verify mentally: 1+4+9=14 matches \frac{3 \cdot 4 \cdot 7}{6}.
Summation pegs
SumClosed form
\sum_{k=1}^{n} k^2\frac{n(n+1)(2n+1)}{6}
\sum_{k=1}^{n} k^3\left(\frac{n(n+1)}{2}\right)^2
First n odd numbersn^2
First n even numbersn(n+1)

Sum of squares

Find \sum_{k=1}^{5} k^2.

  • n=5n=5
  • \sum_{k=1}^{5} k^2 = \frac{5 \cdot 6 \cdot 11}{6}\frac{330}{6}
  • Sum55

Pro tip. The sum of the first n odd numbers is n^2 and of the first n even numbers is n(n+1) — faster than expanding when the sequence is 1+3+5+\ldots or 2+4+6+\ldots.

The sum of the first 7 odd numbers is
  1. 49
  2. 42
  3. 56

First n odds sum to n^2, so 7^2 = 49. Direct addition 1+3+5+7+9+11+13 confirms.

7Telescoping sums

A telescoping sum rewrites each term as a difference so consecutive pieces cancel, leaving only boundary terms. The classic split is \frac{1}{k(k+1)} = \frac1k - \frac{1}{k+1}.

Spot products of consecutive integers in the denominator — partial fractions almost always collapse the middle. After cancellation, only the first positive piece and the last negative piece survive.

Figure. Partial fractions that write each term as a difference leave a chain where every interior summand cancels; only the first positive piece and the last negative piece survive.

How it works

  1. Split each termRewrite \frac{1}{k(k+1)} as \frac1k - \frac{1}{k+1}; analogous splits work for k(k+2) and similar patterns.
  2. Write the expanded sumList enough terms to see \frac12, \frac13, etc. cancelling in pairs.
  3. Keep endpointsAfter cancellation, only the first and last uncancelled pieces remain.

Telescoping fraction sum

Evaluate \sum_{k=1}^{10} \frac{1}{k(k+1)}.

  • Split: \frac{1}{k(k+1)} = \frac1k - \frac{1}{k+1}telescoping form
  • Sum collapses to 1 - \frac{1}{11}1 - \frac{1}{11}
  • Value\frac{10}{11}

Pro tip. Spot the partial-fraction split for products of consecutive terms; telescoping leaves only the first and last pieces.

\sum_{k=1}^{5} \frac{1}{k(k+1)} equals
  1. \frac{5}{6}
  2. \frac{4}{5}
  3. \frac{1}{6}

Telescoping gives 1 - \frac{1}{6} = \frac{5}{6}. \frac{4}{5} is 1 - \frac{1}{5} — off-by-one on the upper limit.

Notes

  • Arithmetic progression: consecutive terms differ by a constant d, and the n-th term is a+(n-1)d.
  • Geometric progression: consecutive terms share a constant ratio r, and the n-th term is ar^{n-1}.
  • Sum of an AP: it equals the number of terms times the average of the first and last term, \frac{n}{2}(a+l).
  • Infinite GP: it converges only when |r|<1, in which case the sum is \frac{a}{1-r}.
  • Harmonic progression: the reciprocals of its terms form an AP, so any HP problem is solved by inverting to an AP first.

Formulas

  • AP n-th term: t_n=a+(n-1)d
  • AP sum: S_n=\frac{n}{2}\big[2a+(n-1)d\big]
  • GP sum: S_n=a\frac{r^n-1}{r-1}
  • Infinite GP: S_\infty=\frac{a}{1-r},\ |r|<1
  • Sum of squares: \sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}
  • Sum of cubes: \sum_{k=1}^{n}k^3=\left(\frac{n(n+1)}{2}\right)^2

Exam traps & shortcuts

  • Any term of an AP is the average of two terms equidistant from it, which pins down a missing middle term at once.
  • For a telescoping sum like \sum\frac{1}{k(k+1)}, split into \frac1k-\frac1{k+1} so nearly every term cancels.
  • The sum of the first n odd numbers is n^2 and of the first n even numbers is n(n+1).
  • When an AP's terms are symmetric, write them as a-d,\ a,\ a+d so the common difference cancels in the sum.

Reference tables

Every row should be reachable from the concept above it — these are pegs, not substitutes for the derivation.

Progression formula sheet
Progressionnth termSum
APt_n = a + (n-1)dS_n = \frac{n}{2}[2a+(n-1)d] = \frac{n}{2}(a+l)
GP (finite)t_n = ar^{n-1}S_n = a\frac{r^n-1}{r-1}
GP (infinite, |r|<1)—S_\infty = \frac{a}{1-r}
HPInvert to AP, then invert back—
\sum k^2—\frac{n(n+1)(2n+1)}{6}
\sum k^3—\left(\frac{n(n+1)}{2}\right)^2

Recap

Read only this the night before.

AP nth term
t_n = a + (n-1)d. A term halfway between two known ones is their average — no need to find d first.
AP sum
n/2 times the average of first and last. If both ends are on the page, skip the expanded bracket.
GP
t_n = ar^{n-1}; finite sum a(r^n-1)/(r-1). Read r once from any consecutive pair.
Infinite GP
Finite sum only when |r|<1: S_\infty = a/(1-r). Verify the ratio before dividing.
HP
Reciprocals form an AP. Invert → solve AP → invert the answer.
Summation pegs
Sum of n odds = n^2; sum of n evens = n(n+1). Squares and cubes have the standard closed forms in the table.
Telescoping
1/[k(k+1)] = 1/k - 1/(k+1). Middle terms cancel; keep the endpoints.

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