Before you begin
How to use this chapter
Sequence questions are rarely about remembering one isolated formula. They test whether you can identify a constant difference, a constant ratio, a cancellation pattern, or a useful relation between a term and its partial sum.
Learn
Read Units 01–07 in order. Try each checkpoint before opening the solution.
Transfer
Use the worked lab to practise choosing a method without being told the topic.
Benchmark
Attempt the named past-paper adaptations, then use each official source for exact timed practice.
You only need algebraic manipulation, factorisation, quadratic equations and basic logarithms.
Language and notation
Sequences, series and partial sums
1.1 A sequence is an ordered function
A sequence \(\{a_n\}\) assigns a term to every allowed positive integer \(n\). Order matters: \(2,5,8,\ldots\) and \(5,2,8,\ldots\) contain similar numbers but are different sequences. A finite sequence stops; an infinite sequence continues.
The value at position \(n\).
The indicated addition of sequence terms.
The sum of the first \(n\) terms.
1.2 The most important bridge: term from sum
Partial-sum difference
Do not apply \(S_n-S_{n-1}\) at \(n=1\) unless \(S_0=0\) is explicitly adopted.
Worked 1 If \(S_n=4n^2-n\), find \(a_n\).
For \(n\ge2\),
\[ a_n=(4n^2-n)-\{4(n-1)^2-(n-1)\}=8n-5. \]The same expression gives \(a_1=3=S_1\), so \(a_n=8n-5\) for every \(n\ge1\).
Method cue: whenever \(S_n\) is given, difference it before trying to identify a progression.
1.3 Sigma notation without index errors
Linearity
\[\sum_{k=1}^{n}(\alpha u_k+\beta v_k)=\alpha\sum_{k=1}^{n}u_k+\beta\sum_{k=1}^{n}v_k\]Constant
\[\sum_{k=1}^{n}c=nc\]Index shift
\[\sum_{k=1}^{n}f(k+1)=\sum_{j=2}^{n+1}f(j)\]Split a range
\[\sum_{k=1}^{n}u_k=\sum_{k=1}^{m}u_k+\sum_{k=m+1}^{n}u_k\]Worked 2 Rewrite \(\sum_{k=3}^{n}(2k-1)\) using a standard sum.
The identity \(1+3+\cdots +(2n-1)=n^2\) makes the subtraction immediate.
Checking the first four terms can suggest a formula; induction, algebra or a recurrence must justify it.
Constant first difference
Arithmetic progressions
An AP has \(a_{n+1}-a_n=d\), independent of \(n\). Starting from first term \(a\), every move adds the same common difference \(d\).
\(n\)th term
\[a_n=a+(n-1)d\]Sum of \(n\) terms
\[S_n=\frac n2\{2a+(n-1)d\}=\frac n2(a+\ell)\]General index gap
\[a_q-a_p=(q-p)d\]Equidistant terms
\[a_{m-r}+a_{m+r}=2a_m\]Deepen the idea
Arithmetic-indexed subsequences of an AP
If \((a_n)\) has common difference \(d\), then \(a_p,a_{p+q},a_{p+2q},\ldots\) is itself an AP, because consecutive selected terms differ by \(qd\). Its first term is \(a_p\). This converts a sum over equally spaced indices into an ordinary AP sum—provided the number of selected indices is counted correctly.
Challenge An AP has \(a_5=11\), \(a_{17}=47\). Find \(a_2+a_5+\cdots+a_{50}\).
Plan. Determine the original AP, then treat the selected terms as a new AP.
Since \(12d=47-11\), \(d=3\), and \(a=-1\). Thus \(a_2=2\), \(a_{50}=146\), and the index list \(2,5,\ldots,50\) contains \((50-2)/3+1=17\) terms.
\[\text{Required sum}=\frac{17}{2}(2+146)=1258.\]Method cue: equally spaced indices multiply the original common difference.
2.1 Structural properties worth using
- Three consecutive AP terms: write \(x-d,x,x+d\), so their sum is \(3x\).
- Four symmetric terms: write \(x-3t,x-t,x+t,x+3t\); their common difference is \(2t\).
- Equal index sums: if \(p+q=r+s\), then \(a_p+a_q=a_r+a_s\).
- Average: the mean of a finite AP is \((\text{first}+\text{last})/2\).
- Odd number of terms: the mean equals the middle term.
Worked 3 The 7th and 19th terms are 23 and 71. Find \(a_{40}\).
Worked 4 An AP has 31 terms and sum 620. Find its 16th term.
With an odd number of terms, the middle term equals the average:
\[ a_{16}=\frac{S_{31}}{31}=\frac{620}{31}=20. \]Fast route: no first term or common difference is needed.
2.2 Inserting arithmetic means
To insert \(m\) arithmetic means between \(x\) and \(y\), build an AP of \(m+2\) terms. There are \(m+1\) equal gaps, so
Worked 5 Insert five arithmetic means between 8 and 50.
The complete AP is \(8,15,22,29,36,43,50\), so the inserted means are \(15,22,29,36,43\).
2.3 Modelling an AP question
Constant multiplicative change
Geometric progressions
A geometric progression is defined by the recurrence \(a_{n+1}=ra_n\), where \(r\) is constant. When every term involved is non-zero this is equivalent to \(a_{n+1}/a_n=r\). The ratio may be positive, negative or zero; a negative ratio produces alternating signs.
\(n\)th term
\[a_n=ar^{n-1}\]Finite sum, \(r\ne1\)
\[S_n=\frac{a(1-r^n)}{1-r}=\frac{a(r^n-1)}{r-1}\]Finite sum, \(r=1\)
\[S_n=na\]Infinite sum
\[S_\infty=\frac a{1-r}\quad\text{only when }|r|<1\]3.1 Why the finite-sum formula works
For \(S_n=a+ar+\cdots+ar^{n-1}\), subtract \(rS_n\) from \(S_n\):
\[ S_n-rS_n=a-ar^n\Rightarrow S_n=\frac{a(1-r^n)}{1-r}. \]The same manoeuvre—multiply by the ratio and subtract—will later handle arithmetico-geometric sums.
Worked 6 Find the sum \(3+6+12+\cdots+1536\).
Here \(a=3,r=2\). Since \(3\cdot2^{n-1}=1536=3\cdot512\), \(n=10\).
\[ S_{10}=3\frac{2^{10}-1}{2-1}=3069. \]3.2 Symmetry and the middle term
- If \(p+q=r+s\), then \(a_pa_q=a_ra_s\).
- For three consecutive non-zero GP terms \(x,y,z\), \(y^2=xz\).
- For an odd number of positive GP terms, the geometric mean of all terms is the middle term.
- Product: for a positive \(n\)-term GP, \(\prod_{k=1}^{n}a_k=(a_1a_n)^{n/2}\); without positivity, use \(\left(\prod_{k=1}^{n}a_k\right)^2=(a_1a_n)^n\).
Worked 7 Three positive GP terms have sum 14 and product 64. Find them.
Write the terms as \(x/r,x,xr\). Their product is \(x^3=64\), hence \(x=4\).
\[ \frac4r+4+4r=14\Rightarrow 2r^2-5r+2=0. \]Thus \(r=2\) or \(1/2\), giving the same ordered set \(2,4,8\) up to reversal.
3.3 Infinite GP: convergence before calculation
An infinite sum exists only if the partial sums approach a finite limit. For a non-zero GP, this requires \(r^n\to0\), exactly when \(|r|<1\).
Worked 8 Evaluate \(12-6+3-\frac32+\cdots\).
The ratio is \(-1/2\), so \(|r|<1\).
\[ S_\infty=\frac{12}{1-(-1/2)}=8. \]3.4 Inserting geometric means
To insert \(m\) real positive geometric means between positive \(x\) and \(y\), use \(xr^{m+1}=y\):
Even roots may eliminate real negative-ratio possibilities. Never take a principal root before checking the sign and parity constraints.
Challenge In a real GP, the first three terms sum to 26 and the next three sum to 702. Find \(a_8\).
The second three-term block is \(r^3\) times the first, so \(26r^3=702\) and \(r=3\). Then \(a(1+r+r^2)=13a=26\), hence \(a=2\).
\[a_8=ar^7=2\cdot3^7=4374.\]Method cue: equal-length consecutive GP blocks differ by a fixed power of the ratio.
Balance and bounds
Arithmetic, geometric and harmonic means
Arithmetic mean
\[A=\frac{x+y}{2}\]Geometric mean
\[G=\sqrt{xy}\quad(x,y>0)\]Harmonic mean
\[H=\frac{2xy}{x+y}\]with equality throughout exactly when \(x=y\).
4.1 Proof of AM–GM for two positive numbers
\[ (\sqrt{x}-\sqrt{y})^2\ge0 \Rightarrow x+y\ge2\sqrt{xy} \Rightarrow \frac{x+y}{2}\ge\sqrt{xy}. \]The equality condition is part of the result. In optimisation, an upper or lower bound is incomplete until you show the equality case is allowed.
Worked 9 Positive \(x,y\) satisfy \(x+y=20\). Find the maximum of \(xy\).
Equality occurs at \(x=y=10\), so the maximum is \(100\).
Worked 10 If the AM and GM of two positive numbers are 10 and 8, find the numbers.
The numbers have sum \(20\) and product \(64\), so they are roots of \(t^2-20t+64=0\).
\[ (t-4)(t-16)=0. \]The two numbers are \(4\) and \(16\).
Challenge Positive \(x,y,z\) satisfy \(x+y+z=12\). Maximise \(x^2yz\).
Apply four-variable AM–GM to \(x/2,x/2,y,z\). Their sum is 12, so their arithmetic mean is 3:
\[\sqrt[4]{\frac{x^2yz}{4}}\le3\quad\Rightarrow\quad x^2yz\le324.\]Equality requires \(x/2=y=z=3\), so \((x,y,z)=(6,3,3)\) and the bound is attainable.
Method cue: repeat a variable according to its target exponent.
4.2 Progression tests for three terms
For positive \(x_i\) and non-negative weights \(\lambda_i\) summing to \(1\), \(\sum \lambda_i x_i\ge\prod x_i^{\lambda_i}\). Use it only when the required exponents match the weights naturally.
High-frequency identities
Standard sums and polynomial terms
5.1 Expand first, then use the smallest formula set
Worked 11 Evaluate \(\sum_{k=1}^{n}k(k+1)\).
Worked 12 Find \(\sum_{k=1}^{12}(3k-1)^2\).
5.2 Difference degree rule
If \(S_n\) is a polynomial of degree \(m\), then \(a_n=S_n-S_{n-1}\) is usually a polynomial of degree \(m-1\). Conversely, summing a polynomial sequence usually raises its degree by one.
A constant second difference signals a quadratic term formula.
Extension · beyond the named syllabus
Telescoping sums and products
A telescoping expression is engineered as \(F(k)-F(k+1)\) or \(F(k)/F(k+1)\). Write the first three and last two expanded terms; the surviving boundary terms become visible.
Telescoping is not separately named in the current syllabus. It is included because it is a reusable algebraic method for integrated JEE problems, not counted as a core completion requirement.
6.1 Partial-fraction templates
Worked 13 Evaluate \(\sum_{k=1}^{n}\frac1{k(k+1)}\).
Worked 14 Evaluate \(\sum_{k=1}^{n}\frac1{(2k-1)(2k+1)}\).
Split each term as a difference of consecutive odd reciprocals:
\[ \frac12\sum_{k=1}^{n}\left(\frac1{2k-1}-\frac1{2k+1}\right) =\frac12\left(1-\frac1{2n+1}\right)=\frac n{2n+1}. \]6.2 Telescoping products
Worked 15 Evaluate \(\prod_{k=2}^{n}\left(1-\frac1{k^2}\right)\).
Most telescope errors are not algebra errors; they are wrong first or last indices. Display the boundary terms before cancelling.
Problem-solving extension · beyond the named syllabus
Arithmetico-geometric sums and recurrences
Deepen the idea
Shift an affine recurrence to its fixed point
For \(a_{n+1}=ra_n+s\), \(r\ne1\), first solve \(L=rL+s\), so \(L=s/(1-r)\). With \(b_n=a_n-L\), the constant disappears:
\[b_{n+1}=rb_n,\qquad a_n=L+(a_1-L)r^{n-1}.\]Thus the shifted sequence is a GP. When \(|r|<1\), the original sequence approaches the fixed point \(L\).
7.1 Arithmetico-geometric progression (AGP)
An AGP multiplies an AP-like coefficient by a GP-like power: \(a+(a+d)r+(a+2d)r^2+\cdots\). Multiply the sum by \(r\), shift, and subtract.
Derivative-of-GP identity
\[ \sum_{k=1}^{n}kr^{k-1} =\frac{1-(n+1)r^n+nr^{n+1}}{(1-r)^2}\quad(r\ne1) \] \[ \sum_{k=1}^{\infty}kr^{k-1}=\frac1{(1-r)^2}\quad(|r|<1). \]Worked 16 Evaluate \(\sum_{k=1}^{\infty}\frac{k}{2^k}\).
Put \(r=1/2\) in \(\sum kr^{k-1}=1/(1-r)^2\), then multiply by \(r\):
\[ \sum_{k=1}^{\infty}\frac{k}{2^k} =\frac{1/2}{(1-1/2)^2}=2. \]7.2 Recurrences become sums
If \(T_{n+1}-T_n=f(n)\), sum both sides from \(n=1\) to \(m-1\):
Worked 17 \(T_1=4\) and \(T_{n+1}-T_n=5n-2\). Find \(T_{20}\).
7.3 Functional equations can generate a GP
If \(f(x+y)=f(x)f(y)\) and \(a_i=a_1+(i-1)d\), then \(f(a_i)=f(a_1)\{f(d)\}^{i-1}\). Thus an AP fed into a multiplicative function becomes a GP.
7.4 The strategy ladder
Challenge Evaluate \(1+\frac23+\frac3{3^2}+\cdots+\frac{10}{3^9}\).
This is \(\sum_{k=1}^{10}kr^{k-1}\) with \(r=1/3\). Apply the finite derivative-of-GP identity:
\[\frac{1-11/3^{10}+10/3^{11}}{(1-1/3)^2}=\frac{44281}{19683}.\]Method cue: a linear coefficient multiplying powers signals an AGP.
Challenge \(a_1=7\), \(a_{n+1}=3a_n-4\). Evaluate \(\sum_{k=1}^{10}1/(a_k-2)\).
The fixed point is 2. Put \(b_n=a_n-2\); then \(b_{n+1}=3b_n\), \(b_1=5\), so \(b_n=5\cdot3^{n-1}\).
\[\sum_{k=1}^{10}\frac1{a_k-2}=\frac15\sum_{j=0}^{9}\left(\frac13\right)^j=\frac3{10}(1-3^{-10}).\]Method cue: subtract the equilibrium before iterating an affine recurrence.
Challenge \(a_1=1\), \(a_{n+1}=a_n/(1+a_n)\). Find \(\prod_{k=1}^{10}(1+a_k)\).
Reciprocals give \(1/a_{n+1}=1/a_n+1\), so \(a_n=1/n\). The recurrence also gives \(1+a_k=a_k/a_{k+1}\).
\[\prod_{k=1}^{10}(1+a_k)=\prod_{k=1}^{10}\frac{a_k}{a_{k+1}}=\frac{a_1}{a_{11}}=11.\]Method cue: a reciprocal substitution can reveal both an AP and a telescoping product.
Classify: difference, ratio, partial sum, or cancellation?
Parameterise: use symmetric terms when sums/products are given.
Compress: use an invariant before solving for every variable.
Check domain: positivity, integrality, convergence and zero denominators.
Verify: substitute into the original condition, not only a transformed equation.
Using \(S_\infty=a/(1-r)\) without \(|r|<1\).
Calling \(a,b,c\) a GP from \(b^2=ac\) when zeros/signs break a common ratio.
Forgetting that \(m\) inserted means create \(m+1\) gaps.
Confusing the \(n\)th term \(a_n\) with the partial sum \(S_n\).
Solving for \(r^2\) and silently discarding the negative ratio.
Cancelling a telescope without writing its final surviving denominator.
Mixed-method training
Worked problem lab
Try to name the method before touching the algebra. Each solution includes the decisive observation.
Worked 18 If the 5th term of an AP is 17 and the 12th is 45, find \(S_{20}\).
Decisive observation: subtract the term equations.
\[ 7d=28\Rightarrow d=4,\qquad a+4d=17\Rightarrow a=1. \] \[ S_{20}=10\{2+19(4)\}=780. \]Worked 19 A positive GP has 2nd term 6 and 5th term 162. Find \(S_6\).
Decisive observation: divide the term conditions.
\[ \frac{ar^4}{ar}=r^3=27\Rightarrow r=3,\quad a=2. \] \[ S_6=2\frac{3^6-1}{3-1}=728. \]Worked 20 The sum of the first \(n\) terms is \(2n^2+3n\). Show that the terms form an AP and find \(S_{25}\).
which is linear in \(n\), hence an AP with \(a_1=5,d=4\). The requested sum is already given:
\[ S_{25}=2(25)^2+3(25)=1325. \]Worked 21 How many 3-term AP subsets can be chosen from \(\{1,2,\ldots,20\}\)?
Decisive observation: fix the positive common difference \(d\).
For \(d=1,\ldots,9\), the first term has \(20-2d\) choices. Therefore
\[ \sum_{d=1}^{9}(20-2d)=180-90=90. \]Worked 22 Solve \(1+2x+3x^2+\cdots=12\) for real \(x\) with \(|x|<1\).
Hence \(1-x=\pm1/\sqrt{12}\), giving \(x=1\mp1/(2\sqrt3)\). Only \(x=1-\frac1{2\sqrt3}\) satisfies \(|x|<1\).
Worked 23 In an AP, \(a_p=q\) and \(a_q=p\), where \(p\ne q\). Find \(a_{p+q}\).
From \(a+(p-1)(-1)=q\), \(a=p+q-1\). Thus
\[ a_{p+q}=p+q-1-(p+q-1)=0. \]Worked 24 A sequence has \(a_1=2,a_2=7\) and constant second difference 6. Find \(a_{10}\).
The first differences form the AP \(5,11,17,\ldots\). Hence
\[ a_{10}=2+\sum_{j=1}^{9}\{5+6(j-1)\} =2+45+216=263. \]Past-paper evidence
Past-paper adaptations and source links
Each item below is an independently edited adaptation of a named official-paper question. Wording, notation or options may be abridged; open the linked PDF for the exact source and exam layout.
Eight named past-paper tasks
Direct structure recognition · target 2–4 minutes each
M1 Intersection of two finite APsJEE Main 2026 · 2 April Shift 1 · Q5 Main
\(A\) contains the first 101 terms of the AP \(1,6,11,\ldots\), and \(B\) contains the first 71 terms of \(9,16,23,\ldots\). How many elements of \(A\cap B\) are divisible by 3?
Answer: 5
Common elements satisfy \(x\equiv1\pmod5\) and \(x\equiv2\pmod7\), hence \(x=16+35t\). Bounds give \(0\le t\le13\). Divisibility by 3 gives \(t\equiv1\pmod3\): \(t=1,4,7,10,13\).
M2 An AP constrained by its last termJEE Main 2026 · 4 April Shift 1 · Q7 Main
A 30-term non-negative AP starts at \(10/3\). Its sum equals the cube of its last term. Find the common difference.
Answer: \(\frac5{87}\)
Let the last term be \(\ell\). Then \(15(10/3+\ell)=\ell^3\), so \(\ell^3-15\ell-50=0\). The admissible real root is \(\ell=5\). Thus \(d=(5-10/3)/29=5/87\).
M3 Probability of choosing three dates in APJEE Main 2026 · 4 April Shift 2 · Q21 Main
A three-element subset of the dates in a 31-day month is chosen uniformly at random. If, after arranging its dates increasingly, they form an AP with probability \(a/b\) in lowest terms, find \(a+b\).
Answer: 944
For common difference \(d=1,\ldots,15\), there are \(31-2d\) valid starts. Favourable subsets \(=225\); total \(=\binom{31}{3}=4495\). The reduced probability is \(45/899\), so \(a+b=944\).
M4 Terms recovered from \(S_n\)JEE Main 2026 · 5 April Shift 1 · Q2 Main
If \(S_n=3n^2+5n\) for an AP, find the sum of squares of its first 10 terms.
Answer: 15220
\(a_n=S_n-S_{n-1}=6n+2\). Therefore \(\sum_{n=1}^{10}(6n+2)^2=36(385)+24(55)+40=15220\).
M5 Quadratic roots placed in GPJEE Main 2026 · 5 April Shift 2 · Q1 Main
Roots \(\alpha,\beta\) of \(x^2-x+p=0\) and roots \(\gamma,\delta\) of \(x^2-4x+q=0\), with integer \(p,q\), occur in that order in a GP. Find \(|p+q|\).
Answer: 34
Write the four terms \(a,ar,ar^2,ar^3\). Their pair sums give \(a(1+r)=1\) and \(ar^2(1+r)=4\), so \(r^2=4\). The choice \(r=2\) makes \(p=2/9\), not an integer. For \(r=-2\), \(a=-1\), \(p=-2,q=-32\), hence 34.
M6 A hidden telescoping sumJEE Main 2026 · 5 April Shift 2 · Q5 Main
The first 10 terms of \(\sum_{k\ge1}\frac{k}{1+4k^4}\) sum to \(m/n\) in lowest terms. Find \(m+n\).
Answer: 276
Use the decomposition in Unit 06. The sum is \(\frac14(1-\frac1{221})=\frac{55}{221}\), so \(m+n=276\).
M7 Mean of selected inserted AMsJEE Main 2026 · 5 April Shift 2 · Q6 Main
Thirty-nine arithmetic means \(A_1,\ldots,A_{39}\) are inserted between 59 and 159. Find the mean of \(A_{25},A_{28},A_{31},A_{36}\).
Answer: 134
The common difference is \(100/40=5/2\). Since an AP is linear in its index, the mean of these terms is \(A_{\text{mean index}}=A_{30}=59+30(5/2)=134\).
M8 Coupled AP and GP conditionsJEE Main 2026 · 6 April Shift 1 · Q6 Main
An AP has first term \(a\), difference \(d\), and \(S_{10}=160\). A GP has first term \(d\), ratio \(a\), and its first two terms sum to 8. Find the sum of all possible values of \(d\).
Answer: \(\frac{34}{9}\)
\(2a+9d=32\) and \(d(1+a)=8\). Eliminating \(d\) gives \(a^2-15a+20=0\). For its roots \(a_1,a_2\),
\[ \sum d=8\left(\frac1{1+a_1}+\frac1{1+a_2}\right) =8\frac{2+(a_1+a_2)}{1+(a_1+a_2)+a_1a_2}=\frac{34}{9}. \]Six named past-paper tasks
Multi-step structure · target 5–10 minutes each
A1 Areas from two APsJEE Advanced 2022 · Paper 1 · Q6 Advanced
Lengths \(l_i\) and widths \(w_i\) are 100-term APs with differences \(d_1,d_2\) and \(d_1d_2=10\). With \(A_i=l_iw_i\), \(A_{51}-A_{50}=1000\). Find \(A_{100}-A_{90}\).
Answer: 18900
\(\Delta_i=A_{i+1}-A_i=C+(2i-1)d_1d_2\). From \(\Delta_{50}=1000\), \(C=10\), so \(\Delta_i=20i\). Hence \(A_{100}-A_{90}=\sum_{i=90}^{99}20i=18900\).
A2 AP-driven recurrenceJEE Advanced 2022 · Paper 1 · Q10 Advanced
\(a_n\) is an AP with \(a_1=7,d=8\). Also \(T_1=3\) and \(T_{n+1}-T_n=a_n\). Find the ordered pair \(\left(T_{30},\sum_{k=1}^{20}T_k\right)\).
Answer: \((3454,10510)\)
\(a_n=8n-1\), hence \(T_n=3+\sum_{j=1}^{n-1}(8j-1)=4n^2-5n+4\). This gives \(T_{30}=3454\) and \(\sum_{k=1}^{20}T_k=10510\).
A3 Integer AP versus integer GPJEE Advanced 2020 · Paper 1 · Q14 Advanced
A positive-integer AP of difference 2 and a positive-integer GP of ratio 2 both start with \(c\). For some positive integer \(n\), twice the AP sum through \(n\) equals the GP sum. How many values of \(c\) are possible?
Answer: 1
The equation reduces to \(c=\frac{2n(n-1)}{2^n-1-2n}\). Testing the only viable small \(n\) gives \(n=3,c=12\); \(n=4,5,6\) are non-integral, and for \(n\ge7\) the fraction lies between 0 and 1.
A4 AP sides in an obtuse triangle: areaJEE Advanced 2023 · Paper 2 · Q14 Advanced
An obtuse triangle has sides in AP, circumradius 1, and largest angle minus smallest angle \(\pi/2\). If its area is \(K\), find \((64K)^2\).
Answer: 1008
Let smallest angle be \(A\); the largest is \(A+\pi/2\), and the middle angle is \(\pi/2-2A\). Since sides and therefore sines are in AP, \(\sin A+\cos A=2\cos2A\), giving \(\cos A-\sin A=1/2\). The sides are \((\sqrt7-1)/2,\sqrt7/2,(\sqrt7+1)/2\). Thus \(K=abc/(4R)=3\sqrt7/16\), and \((64K)^2=1008\).
A5 AP sides in an obtuse triangle: inradiusJEE Advanced 2023 · Paper 2 · Q15 Advanced
An obtuse triangle has sides in AP, circumradius 1, and largest angle minus smallest angle \(\pi/2\). Find its inradius.
Answer: \(0.25\)
The semiperimeter is \(s=3\sqrt7/4\), and the area is \(3\sqrt7/16\). Therefore \(r=(3\sqrt7/16)/(3\sqrt7/4)=1/4=0.25\).
A6 A multiplicative function turns an AP into a GPJEE Advanced 2025 · Paper 1 · Q12 Advanced
A function \(f:\mathbb R\to(0,\infty)\) satisfies \(f(x+y)=f(x)f(y)\). For real AP terms \(a_1,\ldots,a_{50}\), \(f(a_{31})=64f(a_{25})\) and \(\sum_{i=1}^{50}f(a_i)=3(2^{25}+1)\). Find \(\sum_{i=6}^{30}f(a_i)\).
Answer: 96
Let the AP difference be \(d\). Then \(f(a_i)=c\,q^{i-1}\), where \(q=f(d)>0\). The ratio condition gives \(q^6=64\), hence \(q=2\). From \(c(2^{50}-1)=3(2^{25}+1)\), \(c=3/(2^{25}-1)\). Therefore the requested 25-term sum is \(c\,2^5(2^{25}-1)=96\).
Timed application
Practice tests
These are original RankDeterminant tests. Your score is computed locally in the page; unanswered questions receive zero and there is no negative marking. Use the timer as a pacing tool, not a rigid rule.
Complete the test, then check your work.
Exact-match scoring is used for multi-select items.
Last-day review
One-page revision sheet
AP
\(a_n=a+(n-1)d\)
\(S_n=\frac n2[2a+(n-1)d]=\frac n2(a+\ell)\)
\(a_p+a_q=a_r+a_s\) if \(p+q=r+s\)
\(m\) inserted AMs: \(d=(y-x)/(m+1)\)
GP
\(a_n=ar^{n-1}\)
\(S_n=a(1-r^n)/(1-r)\), \(r\ne1\)
\(S_\infty=a/(1-r)\), only \(|r|<1\)
\(a_pa_q=a_ra_s\) if \(p+q=r+s\)
Means
\(A=(x+y)/2,\;G=\sqrt{xy},\;H=2xy/(x+y)\)
\(A\ge G\ge H,\; AH=G^2\)
AP: \(2b=a+c\) · GP: \(b^2=ac\)
Equality in AM–GM requires equal positive terms.
Standard sums
\(\sum k=n(n+1)/2\)
\(\sum k^2=n(n+1)(2n+1)/6\)
\(\sum k^3=[n(n+1)/2]^2\)
\(\sum kr^{k-1}=\frac{1-(n+1)r^n+nr^{n+1}}{(1-r)^2}\)
First moves
Given \(S_n\) → compute \(S_n-S_{n-1}\).
Rational summand → factor denominator.
Symmetric AP/GP terms → parameterise around the middle.
Recurrence difference → sum both sides.
Non-negotiable checks
Convergence: \(|r|<1\).
Index bounds and number of gaps.
Sign and integer restrictions.
Substitute candidate answers back.
Mastery checklist
Primary references
Syllabus and paper sources
Scope and PYQ metadata were checked against official NTA and JEE Advanced material. RankDeterminant is an independent learning resource and is not affiliated with NTA, IITs or JoSAA.