JEE Advanced (IIT Entrance) · Mathematics (JEE & NDA)
Sequence and Series
Arithmetic and geometric progressions, arithmetic and geometric means, and the relation between the two means.
Six concepts. Progressions give the term and the sum; means sit between two positives; power sums close polynomial series; AG and telescoping finish the paper tricks. Every ledger below is a derivation, not a recomputation of rupees.
- JEE Advanced (IIT Entrance)
- Medium level
- 6 concepts
- 5 practice questions
1Arithmetic progression: term and sum
An arithmetic progression has a constant common difference d. The n-th term is a_n=a+(n-1)d. The sum of the first n terms is S_n=\dfrac{n}{2}[2a+(n-1)d], or equivalently S_n=\dfrac{n}{2}(a+l) when the last term l is already in hand — both forms are the same identity written two ways.
Figure. An AP is equal steps on the number line. Here a=3 and each hop is the common difference d=4, matching the worked example.
How it works
- Read a and dFirst term a, common difference from any consecutive pair. Sign of d is the direction of the progression.
- Write a_n or lNeed a specific term: a_n=a+(n-1)d. Already know the last term: call it l and skip straight to the second sum form.
- SumS_n=\dfrac{n}{2}[2a+(n-1)d] or S_n=\dfrac{n}{2}(a+l). Same answer; pick the form that uses what you already have.
AP sum
Find the sum of the first 20 terms of the AP 3,7,11,\dots.
- a=3, d=4, n=20setup
- S_{20}=\dfrac{20}{2}[2(3)+(20-1)(4)]10[6+76]
- 10\times 82820
Pro tip. Three terms in AP are cleaner as a-d,\,a,\,a+d: the middle is the mean and the common difference sits in one letter. If the last term is easier than expanding (n-1)d, use S_n=\tfrac{n}{2}(a+l) — both forms give the same result.
The 10th term of the AP 3,7,11,\dots is
- 39
- 43
- 35
a_n=a+(n-1)d=3+9\cdot 4=39. 43 is the 11th term; 35 uses n instead of n-1.
2Geometric progression: term and sum
A geometric progression has a constant common ratio r. The n-th term is a_n=ar^{n-1}. The finite sum is S_n=a\dfrac{r^n-1}{r-1} when r\ne 1. An infinite GP converges only when the absolute value of the common ratio is strictly less than one, and then S_\infty=\dfrac{a}{1-r} — apply that formula only after checking the ratio.
Figure. Each term is the previous times r=1/3. Because |r|<1 the terms shrink to zero and the infinite sum is a/(1-r)=9.
How it works
- Read a and rFirst term a; common ratio from any consecutive pair a_{k+1}/a_k. Watch the sign of r.
- Finite or infinite?Asked for n terms: use S_n. Asked for sum to infinity: first demand |r|<1, then S_\infty=a/(1-r).
- EvaluateSubstitute carefully — a negative r alternates, and r=1 is the constant sequence a+a+\cdots=na, not the GP sum formula.
Sum of an infinite GP
Find the sum to infinity of the series 6+2+\dfrac{2}{3}+\dfrac{2}{9}+\dots.
- a=6, r=\dfrac{2}{6}=\dfrac{1}{3} with |r|<1converges
- S_\infty=\dfrac{a}{1-r}=\dfrac{6}{1-\tfrac{1}{3}}\dfrac{6}{\tfrac{2}{3}}
- 6\times\dfrac{3}{2}9
Pro tip. Three terms in GP are cleaner as \dfrac{a}{r},\,a,\,ar: the middle is the geometric mean and the product of extremes is a^2. Never write a/(1-r) until |r|<1 is checked — that is the whole condition for convergence, not a tip about speed.
The infinite series 1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dots sums to
- 2
- 1
- diverges
a=1, r=1/2, |r|<1, so S_\infty=1/(1-1/2)=2. Claiming it diverges confuses |r|<1 with |r|>1; summing only the first term gives 1.
3AM, GM, HM and the inequality
Between two positive numbers a and b, the arithmetic mean is A=\dfrac{a+b}{2}, the geometric mean is G=\sqrt{ab}, and the harmonic mean is H=\dfrac{2ab}{a+b}. They always satisfy A\ge G\ge H, with equality if and only if the two numbers are equal, and the identity G^2=AH ties all three together.
Figure. For a=4, b=16 the means fall in order H=6.4, G=8, A=10 between the endpoints — A\ge G\ge H, and G^2=AH.
How it works
- Form the three meansFor positive a,b: A=(a+b)/2, G=\sqrt{ab}, H=2ab/(a+b).
- Order themA\ge G\ge H, equality precisely when a=b. Never reverse the chain for positives.
- Use A\ge G for a boundA minimum often drops out of A\ge G in one line — classic: for x>0, x+1/x\ge 2.
| Mean | Formula | Role |
|---|---|---|
| Arithmetic A | (a+b)/2 | Largest of the three |
| Geometric G | \sqrt{ab} | Middle; G^2=AH |
| Harmonic H | 2ab/(a+b) | Smallest of the three |
Minimum via A\ge G
For x>0, show that x+\dfrac{1}{x}\ge 2 and find when equality holds.
- A\ge G on x and 1/x: \dfrac{x+1/x}{2}\ge\sqrt{x\cdot 1/x}\dfrac{x+1/x}{2}\ge 1
- Multiply by 2x+\dfrac{1}{x}\ge 2
- Equality in A\ge G iff the two numbers matchx=1/x\Rightarrow x=1
Pro tip. Use A\ge G to find minimum values quickly: for positive x, x+1/x\ge 2 with equality at x=1. The threshold answers when the lower bound is attained, not when the expression stops decreasing for other reasons.
For positive a\ne b, which is true?
- A>G>H
- A=G=H
- H>G>A
Strict inequality when a\ne b. Equality of all three holds only for a=b; reversing the order confuses AM with HM.
4Power sums \sum k, \sum k^2, \sum k^3
Three closed forms evaluate most polynomial series on sight: \sum_{k=1}^{n}k=\dfrac{n(n+1)}{2}, \sum_{k=1}^{n}k^2=\dfrac{n(n+1)(2n+1)}{6}, and \sum_{k=1}^{n}k^3=\left[\dfrac{n(n+1)}{2}\right]^2. Expand a polynomial in the index, sum term by term, and substitute — no induction mid-paper.
Closed forms for \sum k, \sum k^2, \sum k^3 are algebraic identities; substitute into them on the page.
| Sum | Closed form |
|---|---|
| \sum_{k=1}^{n} k | n(n+1)/2 |
| \sum_{k=1}^{n} k^2 | n(n+1)(2n+1)/6 |
| \sum_{k=1}^{n} k^3 | [n(n+1)/2]^2 |
\sum_{k=1}^{10} k^2 equals
- 385
- 55
- 3025
n(n+1)(2n+1)/6=10\cdot 11\cdot 21/6=385. 55 is \sum k; 3025 is \sum k^3.
5Arithmetico-geometric series
An arithmetico-geometric series multiplies an AP factor by a GP factor termwise — typical shape a+(a+d)r+(a+2d)r^2+\cdots. The standard attack is the same every time: write the sum, multiply by the common ratio and subtract, and the difference collapses to a GP (plus at most one leftover AP-edge term).
Figure. Write S=a+(a+d)r+(a+2d)r^2+\cdots, form rS shifted one GP place to the right, and subtract. The mixed middle cancels and S-rS is a geometric series.
The S-rS method
- Write SAlign powers of r under the AP coefficients so each column is one term.
- Form rSMultiply every term by r, shifting the GP part one place to the right.
- SubtractS-rS cancels the mixed middle and leaves a pure GP (and possibly a final unpaired term). Sum that GP and solve for S.
6Telescoping series
A telescoping series collapses because consecutive terms cancel after a partial-fraction split. The prototype is \sum\dfrac{1}{k(k+1)}: write \dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}, and the sum from k=1 to n leaves only the first positive piece and the last negative piece.
Figure. After \frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}, the sum from k=1 to 5 is (1-1/2)+(1/2-1/3)+\cdots+(1/5-1/6). Middles cancel and S_5=1-1/6=5/6.
How it works
- SplitDecompose the general term by partial fractions so each summand is a difference of two simpler pieces.
- Write the partial sumExpand S_n=(u_1-u_2)+(u_2-u_3)+\cdots+(u_n-u_{n+1}) and watch the middle cancel.
- CloseWhat remains is u_1-u_{n+1}. Take n\to\infty only if that limit exists.
Telescoping prototype
Evaluate S_n=\sum_{k=1}^{n}\dfrac{1}{k(k+1)}.
- \dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}partial fractions
- S_n=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\cdots+\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)1-\dfrac{1}{n+1}
- Closed formS_n=\dfrac{n}{n+1}
Pro tip. For a telescoping sum, split each term by partial fractions so consecutive pieces cancel, leaving only the first and last. If the remaining last term does not tend to a limit, the infinite series diverges even though cancellation happened.
\sum_{k=1}^{5}\dfrac{1}{k(k+1)} equals
- 5/6
- 1
- 1/5
S_n=n/(n+1), so S_5=5/6. 1 is the n\to\infty limit; 1/5 is a single end term.
Notes
- Arithmetic progression: An AP has common difference d, n-th term a_n=a+(n-1)d and sum S_n=\dfrac{n}{2}[2a+(n-1)d]=\dfrac{n}{2}(a+l), where l is the last term.
- Geometric progression: A GP has common ratio r, n-th term a_n=ar^{n-1} and sum S_n=a\dfrac{r^n-1}{r-1} (r\ne1). An infinite GP with |r|<1 sums to S_\infty=\dfrac{a}{1-r}.
- Means and their inequality: Between two positive numbers, the arithmetic mean A=\dfrac{a+b}{2}, geometric mean G=\sqrt{ab} and harmonic mean H=\dfrac{2ab}{a+b} satisfy A\ge G\ge H with G^2=AH.
- Special sums: \sum_{k=1}^{n}k=\dfrac{n(n+1)}{2}, \sum k^2=\dfrac{n(n+1)(2n+1)}{6} and \sum k^3=\left[\dfrac{n(n+1)}{2}\right]^2. These evaluate many polynomial series directly.
- Arithmetico-geometric and telescoping series: Products of an AP and a GP are summed by multiplying by r and subtracting; telescoping series such as \sum\dfrac{1}{k(k+1)} collapse using partial fractions \dfrac{1}{k}-\dfrac{1}{k+1}.
- Method of differences: if the term T_n can be written as V_{n}-V_{n-1}, the series telescopes to V_N-V_0.
- AGP sums (arithmetic times geometric) are summed by the S - rS shift technique, exactly like a geometric series.
- For sums like \sum \dfrac{1}{n(n+1)(n+2)} use partial fractions to telescope.
- AM-GM-HM inequality and Cauchy-Schwarz frequently bound series-type expressions in Advanced problems.
Formulas
- AP: a_n=a+(n-1)d,\quad S_n=\dfrac{n}{2}[2a+(n-1)d]
- GP: a_n=ar^{n-1},\quad S_n=a\dfrac{r^n-1}{r-1},\quad S_\infty=\dfrac{a}{1-r}\ (|r|<1)
- Means: A=\dfrac{a+b}{2},\ G=\sqrt{ab},\ H=\dfrac{2ab}{a+b},\ G^2=AH
- Power sums: \sum k=\dfrac{n(n+1)}{2},\ \sum k^2=\dfrac{n(n+1)(2n+1)}{6}
- Cubes: \sum k^3=\left[\dfrac{n(n+1)}{2}\right]^2
- Inequality: A\ge G\ge H
- \sum_{k=1}^{n} k^3 = \left(\dfrac{n(n+1)}{2}\right)^2
- AGP: \sum_{k=0}^{n-1}(a+kd)r^k summed via S-rS
- \sum_{k=1}^{n}\dfrac{1}{k(k+1)} = 1-\dfrac{1}{n+1}
Exam traps & shortcuts
- Choose symmetric terms to simplify unknowns: three terms in AP as a-d,a,a+d and three in GP as \dfrac{a}{r},a,ar so the middle term and product/sum are clean.
- For a telescoping sum, split each term by partial fractions so consecutive pieces cancel, leaving only the first and last.
- Use A\ge G to find minimum values quickly: for positive x, x+\dfrac{1}{x}\ge2 with equality at x=1.
- Whenever a denominator is a product of consecutive linear factors, decompose into partial fractions to telescope.
- Recognise AM \ge GM to get quick bounds and to identify equality cases in optimisation-flavoured series problems.
Reference tables
The six formulas that close almost every Main question in this chapter.
| Object | Formula | Watch |
|---|---|---|
| AP term | a_n=a+(n-1)d | Index is n-1, not n |
| AP sum | S_n=\dfrac{n}{2}[2a+(n-1)d]=\dfrac{n}{2}(a+l) | Same identity, two shapes |
| GP term | a_n=ar^{n-1} | First power is r^0=1 |
| GP finite sum | S_n=a\dfrac{r^n-1}{r-1} (r\ne 1) | Constant sequence when r=1 |
| GP infinite sum | S_\infty=\dfrac{a}{1-r} | Only if |r|<1 |
| Means | A\ge G\ge H, G^2=AH | Positives; equality iff a=b |
Recap
Read only this the night before.
- AP
- a_n=a+(n-1)d, S_n=n/2\,[2a+(n-1)d]. Symmetric unknowns as a-d,a,a+d.
- GP
- a_n=ar^{n-1}. Infinite sum a/(1-r) only when |r|<1 — check before writing the formula.
- Means
- A\ge G\ge H for positives, G^2=AH, equality iff a=b. Bound minima with A\ge G.
- Power sums
- \sum k, \sum k^2, \sum k^3 are the three closed forms; expand and substitute.
- AG / telescope
- AG: write S, form rS, subtract. Telescope: partial fractions, cancel middle, keep ends.
Practise Sequence and Series
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- 5 exam-style questions on this topic, with explanations
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