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JEE Mains Maths · Formula sheet

Sequences and Series formulas

22 formulas and 22 common traps for JEE Mains Maths Sequences and Series, grouped by subtopic.

Full notes with worked examples

Arithmetic Progressions: Terms and Sums

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The nth term and the common difference

nth term

an=a+(n−1)d,d=aq−apq−pa_n=a+(n-1)d,\qquad d=\frac{a_q-a_p}{q-p}

Sum of n terms, and an AP given by its sum

Sum of n terms

Sn=n2[2a+(n−1)d]S_n=\frac n2\big[2a+(n-1)d\big]

Symmetric terms, equal pairs and odd–even parts

Odd–even parts of 2k terms

Seven−Sodd=kd,ak+an+1−k=a1+anS_{\text{even}}-S_{\text{odd}}=kd,\qquad a_k+a_{n+1-k}=a_1+a_n

Common traps

Count steps, not terms

From apa_p to aqa_q there are q−pq-p steps. With nn means inserted there are n+1n+1 steps, not nn.

A constant term in S_n

If the given SnS_n has a constant term, S1S_1 does not follow the pattern of Sn−Sn−1S_n-S_{n-1}: the sequence is an AP only from the second term.

Odd-placed, not odd-valued

The odd terms of an AP are a1,a3,a5,…a_1,a_3,a_5,\dots, the terms in odd positions. Their values need not be odd numbers.

Common Terms and Sub-Progressions

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Common terms of two progressions

Common difference of the shared terms

D=lcm⁡(d1,d2)D=\operatorname{lcm}(d_1,d_2)

Terms picked out by a divisibility condition

Sum of the terms that pass

sum=Sall−Sexcluded\text{sum}=S_{\text{all}}-S_{\text{excluded}}

Common traps

Start at the first COMMON term

The shared AP starts at the first term both lists contain, which is usually neither progression's first term.

Add the overlap back

Removing multiples of 3 and then multiples of 5 removes the multiples of 15 twice. Add their sum back once.

Geometric Progressions: Terms and Sums

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Terms and sums of a GP from two conditions

Sum of n terms

Sn=a rn−1r−1S_n=a\,\frac{r^n-1}{r-1}

Products, means and odd–even parts of a GP

Whole sum of 2m terms

S=(1+r) SoddS=(1+r)\,S_{\text{odd}}

Recognising a GP in a function, a recurrence or a two-base sum

Two-base sum

∑k=0mxkym−k=ym+1−xm+1y−x\sum_{k=0}^{m}x^{k}y^{m-k}=\frac{y^{m+1}-x^{m+1}}{y-x}

Common traps

Increasing and positive means r > 1

Dividing two conditions often gives a quadratic in rr with roots tt and 1t\frac1t. An increasing GP of positive terms keeps only the root above 1.

An odd number of terms

With an odd count there is one more odd-placed term than even-placed, and the factor 1+r1+r no longer holds. Write both sums out.

f(x + y) = 2f(x)f(y) is not f(1)^n

Put g=2fg=2f: then g(x+y)=g(x)g(y)g(x+y)=g(x)g(y), so g(n)=g(1)ng(n)=g(1)^n and f(n)=12(2f(1))nf(n)=\frac12(2f(1))^n.

AP and GP Conditions, Means and AM–GM

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Turning 'in AP', 'in GP' and 'in HP' into equations

Three conditions

2b=a+c,b2=ac,b=2aca+c2b=a+c,\qquad b^2=ac,\qquad b=\frac{2ac}{a+c}

AM, GM and HM of two numbers, and means inserted between them

The three means

A≥G≥H,G2=A HA\ge G\ge H,\qquad G^2=A\,H

Least and greatest values by AM–GM

AM–GM

x1+x2+⋯+xnn≥(x1x2⋯xn)1/n\frac{x_1+x_2+\dots+x_n}{n}\ge\left(x_1x_2\cdots x_n\right)^{1/n}

Common traps

A zero term is not a GP

A root that makes any GP term zero, or the ratio undefined, must be rejected even though it satisfies b2=acb^2=ac.

The larger root is the AM

When the AM and GM are given as the roots of a quadratic, the larger root is the AM, because A≥GA\ge G.

Check the equality case

AM–GM gives a bound, and the bound is the answer only if the equal-pieces point is allowed: positive, and inside any stated range.

Infinite Geometric Series

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Sum to infinity, and the series of squares

Sum to infinity

S∞=a1−r,∣r∣<1S_\infty=\frac{a}{1-r},\quad |r|<1

Infinite GPs inside exponents, logarithms and figures

The geometric series

1+x+x2+⋯=11−x,∣x∣<11+x+x^2+\dots=\frac{1}{1-x},\quad |x|<1

Common traps

Check |r| < 1

A quadratic in rr may give a root with ∣r∣≥1|r|\ge1. The sum to infinity does not exist there, so that root is rejected.

The ratio must stay below 1

cos⁡2x+cos⁡4x+…\cos^2x+\cos^4x+\dots diverges where cos⁡2x=1\cos^2x=1. Check the stated range keeps the ratio strictly between −1 and 1.

Sums by Standard Formulas

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Σk, Σk² and Σk³ applied to a polynomial kth term

Sum of cubes

∑k=1nk3=[n(n+1)2]2\sum_{k=1}^{n}k^3=\left[\frac{n(n+1)}{2}\right]^2

Finding the kth term from differences or from the sum

Quadratic kth term

Tk=ak2+bk+c,2a=second differenceT_k=ak^2+bk+c,\qquad 2a=\text{second difference}

Alternating, grouped and greatest-integer sums

Alternating squares

12−22+32−⋯−(2n)2=−n(2n+1)1^2-2^2+3^2-\dots-(2n)^2=-n(2n+1)

Common traps

Sum first, then put in the limit

Expand the term in kk, add over kk, and only then put in the upper limit. Putting nn inside the term too early mixes the running index with the limit.

Check the fourth term

Three terms always fit some quadratic. Check the fitted TkT_k against a fourth term before adding; if it fails, the differences are not in AP.

An odd count leaves one term unpaired

Pairing works cleanly only for an even number of terms. With an odd count, pair the rest and add the last term separately.

Telescoping Sums

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Partial fractions that cancel in pairs

Telescoping

∑k=1n[f(k)−f(k+1)]=f(1)−f(n+1)\sum_{k=1}^{n}\big[f(k)-f(k+1)\big]=f(1)-f(n+1)

Quartic denominators and surds

Quartic split

kk4+k2+1=12[1k2−k+1−1k2+k+1]\frac{k}{k^4+k^2+1}=\frac12\left[\frac{1}{k^2-k+1}-\frac{1}{k^2+k+1}\right]

Factorials, powers and functions that telescope

Factorial telescoping

k⋅k!=(k+1)!−k!k\cdot k!=(k+1)!-k!

Common traps

A gap of two leaves two terms at each end

In 12(1k−1k+2)\frac12\left(\frac1k-\frac1{k+2}\right) the first two positive terms and the last two negative terms survive, not one of each.

Keep the ½

The two factors differ by 2k2k, not kk, so the split carries a factor 12\frac12. Dropping it doubles the answer, and the doubled value is usually an option.

The leftover at the lower limit

The sum is Q(n+1)(n+1)!−Q(1)⋅1!Q(n+1)(n+1)!-Q(1)\cdot1!. Q(1)Q(1) is often zero, but not always: work it out before dropping it.

Arithmetico-Geometric and Exponential Series

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Multiply by the ratio and subtract

Infinite AGP

S∞=a1−r+dr(1−r)2S_\infty=\frac{a}{1-r}+\frac{dr}{(1-r)^2}

Numerators whose differences are in AP

Quadratic numerators

∑k≥1k(k+1) xk−1=2(1−x)3\sum_{k\ge1}k(k+1)\,x^{k-1}=\frac{2}{(1-x)^3}

Series summed through e and the logarithm series

The exponential pieces

∑n≥0n(n−1)n!=∑n≥0nn!=e\sum_{n\ge0}\frac{n(n-1)}{n!}=\sum_{n\ge0}\frac{n}{n!}=e

Common traps

The finite case has a last term

For a finite AGP, the subtraction leaves −(a+(n−1)d)rn-\big(a+(n-1)d\big)r^n at the end. Dropping it gives the infinite formula, which is wrong here.

One subtraction is not enough

After the first subtraction the numerators are in AP, so it is still an arithmetico-geometric series. Subtract again before using the GP formula.

Watch where the sum starts

A sum from n=1n=1 leaves out the n=0n=0 term of the ee series. Subtract that term, or the answer is off by a constant.

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