Advanced Sequences, Inequalities, and Straight Lines

Advanced Sequence Differences

  • First Difference in Arithmetic Progression (AP):     * If the first difference of a sequence is in AP, the general term tnt_n takes the quadratic form:     * tn=an2+bn+ct_n = an^2 + bn + c

  • First Difference in Geometric Progression (GP):     * If the first difference is in GP, the general term takes the form:     * tn=a(rn)+bn+ct_n = a(r^n) + bn + c

  • Second Difference in AP:     * If the second difference of a sequence is in AP, the general term is a cubic polynomial:     * tn=an3+bn2+cn+dt_n = an^3 + bn^2 + cn + d

  • Second Difference in GP:     * If the second difference is in GP, the general term takes the form:     * tn=a(rn)+cn2+dn+et_n = a(r^n) + cn^2 + dn + e

  • Case Study Example:     * Sequence: 5+7+13+31+85+5 + 7 + 13 + 31 + 85 + \dots up to 10 terms. This was analyzed in the context of identifying the order of difference.

Telescopic Method of Difference

  • Core Principle:     * Express the general term trt_r as a difference of two terms of a function, such as tn=f(n+1)f(n)t_n = f(n+1) - f(n).     * Summation formula: Sn=r=1ntr=[f(2)f(1)]+[f(3)f(2)]++[f(n+1)f(n)]S_n = \sum_{r=1}^n t_r = [f(2) - f(1)] + [f(3) - f(2)] + \dots + [f(n+1) - f(n)]     * Cancellation occurs, leaving only the first and last terms: Sn=f(n+1)f(1)S_n = f(n+1) - f(1).

  • Application Example 1 (Product of Odds):     * Find the sum of 10 terms of S=(1×3×5)+(3×5×7)+(5×7×9)+S = (1 \times 3 \times 5) + (3 \times 5 \times 7) + (5 \times 7 \times 9) + \dots     * General term: tr=(2r1)(2r+1)(2r+3)t_r = (2r - 1)(2r + 1)(2r + 3).     * Transformation: Multiply and divide by the constant difference of the extended sequence. Use (2r+5)(2r3)=8(2r + 5) - (2r - 3) = 8.     * tr=18[(2r1)(2r+1)(2r+3)(2r+5)(2r3)(2r1)(2r+1)(2r+3)]t_r = \frac{1}{8} [(2r-1)(2r+1)(2r+3)(2r+5) - (2r-3)(2r-1)(2r+1)(2r+3)]     * Final Sum S10=18[f(10)f(0)]=18[(19×21×23×25)+15]S_{10} = \frac{1}{8} [f(10) - f(0)] = \frac{1}{8} [(19 \times 21 \times 23 \times 25) + 15].

  • Application Example 2 (Fractions with Odds):     * Series: Sn=11×3+21×3×5+31×3×5×7+S_n = \frac{1}{1 \times 3} + \frac{2}{1 \times 3 \times 5} + \frac{3}{1 \times 3 \times 5 \times 7} + \dots     * General term: tr=r1×3×5(2r+1)t_r = \frac{r}{1 \times 3 \times 5 \dots (2r+1)}.     * Transformation: Multiply by 2. 2r=(2r+1)12r = (2r + 1) - 1.     * tr=12[(2r+1)11×3(2r+1)]=12[11×3(2r1)11×3(2r+1)]t_r = \frac{1}{2} [\frac{(2r+1) - 1}{1 \times 3 \dots (2r+1)}] = \frac{1}{2} [\frac{1}{1 \times 3 \dots (2r-1)} - \frac{1}{1 \times 3 \dots (2r+1)}].     * Sum Sn=12[111×3(2n+1)]S_n = \frac{1}{2} [1 - \frac{1}{1 \times 3 \dots (2n+1)}].

  • JEE Mains 2021 Question:     * Calculate 1321+1521+1721++120121\frac{1}{3^2-1} + \frac{1}{5^2-1} + \frac{1}{7^2-1} + \dots + \frac{1}{201^2-1}.     * tr=1(2r+1)21=14r(r+1)=14[1r1r+1]t_r = \frac{1}{(2r+1)^2 - 1} = \frac{1}{4r(r+1)} = \frac{1}{4} [\frac{1}{r} - \frac{1}{r+1}].     * Sum for n=100n=100: 14[11101]=25101\frac{1}{4} [1 - \frac{1}{101}] = \frac{25}{101}.

  • JEE Mains 2021 Question 2:     * Sum of 10 terms of 312×22+522×32+732×42+\frac{3}{1^2 \times 2^2} + \frac{5}{2^2 \times 3^2} + \frac{7}{3^2 \times 4^2} + \dots     * General term: tr=2r+1r2(r+1)2=(r+1)2r2r2(r+1)2=1r21(r+1)2t_r = \frac{2r+1}{r^2(r+1)^2} = \frac{(r+1)^2 - r^2}{r^2(r+1)^2} = \frac{1}{r^2} - \frac{1}{(r+1)^2}.     * Result: 11112=1201211 - \frac{1}{11^2} = \frac{120}{121}.

Factorial Series and Summation

  • Product Property:     * n×n!=((n+1)1)n!=(n+1)!n!n \times n! = ((n+1) - 1)n! = (n+1)! - n!

  • Example Case:     * S=11!+22!+33!++100100!S = 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \dots + 100 \cdot 100!     * tr=(r+1)!r!t_r = (r+1)! - r!     * S=101!1!S = 101! - 1!

  • Example Case (Fractional Factorials):     * S=12!+23!+34!++5051!S = \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \dots + \frac{50}{51!}     * tr=r(r+1)!=(r+1)1(r+1)!=1r!1(r+1)!t_r = \frac{r}{(r+1)!} = \frac{(r+1) - 1}{(r+1)!} = \frac{1}{r!} - \frac{1}{(r+1)!}.     * Sum: 1151!1 - \frac{1}{51!}.

Arithmetic, Geometric, Harmonic, and Root Mean Square Inequalities

  • Theorem for Positive Real Numbers:     * For positive numbers a1,a2,,ana_1, a_2, \dots, a_n, the following hierarchy holds:     * RMSAMGMHMRMS \ge AM \ge GM \ge HM

  • Definitions for Two Positive Numbers (a,ba, b):     * Arithmetic Mean (AM): a+b2\frac{a+b}{2}     * Geometric Mean (GM): ab\sqrt{ab}     * Harmonic Mean (HM): 2aba+b\frac{2ab}{a+b}     * Root Mean Square (RMS): a2+b22\sqrt{\frac{a^2+b^2}{2}}

  • General Formulas for $n$ Numbers:     * AM=ainAM = \frac{\sum a_i}{n}     * GM=(a1a2an)1nGM = (a_1 a_2 \dots a_n)^{\frac{1}{n}}     * HM=n1aiHM = \frac{n}{\sum \frac{1}{a_i}}     * RMS=ai2nRMS = \sqrt{\frac{\sum a_i^2}{n}}

  • Equality Condition:     * The equality AM=GM=HM=RMSAM = GM = HM = RMS holds if and only if a1=a2==ana_1 = a_2 = \dots = a_n.

  • Application - Minimum Values:     * For f(x)=x+1xf(x) = x + \frac{1}{x} where x > 0, since AMGMAM \ge GM: x+1x2x1x=1    x+1x2\frac{x + \frac{1}{x}}{2} \ge \sqrt{x \cdot \frac{1}{x}} = 1\implies x + \frac{1}{x} \ge 2. Range: [2,)[2, \infty).     * JEE Advanced 2011: Minimum value of a5+a4+3a3+1+a8+a10a^{-5} + a^{-4} + 3a^{-3} + 1 + a^8 + a^{10} for a > 0.         * Split 3a33a^{-3} into a3+a3+a3a^{-3} + a^{-3} + a^{-3}. There are 8 terms total.         * Check product: (a5)(a4)(a3)3(1)(a8)(a10)=a549+18=a0=1(a^{-5})(a^{-4})(a^{-3})^3(1)(a^8)(a^{10}) = a^{-5-4-9+18} = a^0 = 1.         * AMGM    Sum8(1)18    Minimum Sum=8AM \ge GM \implies \frac{\text{Sum}}{8} \ge (1)^{\frac{1}{8}} \implies \text{Minimum Sum} = 8.

Power Series Expansions and Maclaurin Series

  • Maclaurin Series Formula:     * f(x)=f(0)+f(0)1!x+f(0)2!x2+f(0)3!x3+f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots

  • Exponential Series (exe^x):     * ex=1+x1!+x22!+x33!+=r=0xrr!e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots = \sum_{r=0}^{\infty} \frac{x^r}{r!}     * ex=1x1!+x22!x33!+e^{-x} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots

  • Combined Exponential Sums:     * ex+ex2=1+x22!+x44!+\frac{e^x + e^{-x}}{2} = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots     * exex2=x1!+x33!+x55!+\frac{e^x - e^{-x}}{2} = \frac{x}{1!} + \frac{x^3}{3!} + \frac{x^5}{5!} + \dots

  • Logarithmic Series:     * Valid for -1 < x \le 1: ln(1+x)=xx22+x33x44+\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots     * ln(1x)=xx22x33\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \dots

  • Homework Exercise:     * Evaluate n=0n3n!\sum_{n=0}^{\infty} \frac{n^3}{n!}.

Straight Lines - Fundamentals of Coordinate Geometry

  • Distances from Axes:     * Distance of point (x,y)(x, y) from X-axis = y|y| (Modulus of Ordinate).     * Distance of point (x,y)(x, y) from Y-axis = x|x| (Modulus of Abscissa).

  • Relative Movement:     * Moving Right/Left: y-coordinate reflects no change; x-coordinate increases/decreases.     * Moving Up/Down: x-coordinate reflects no change; y-coordinate increases/decreases.

  • Quadrant Logic (JEE Advanced 2011 Example):     * Point coordinates involving logarithms: Determining signs of terms like 5+log2(x)5 + \log_2(x) to identify the quadrant.

  • Section Formula:     * Internal Division: P(x,y)=(mx2+nx1m+n,my2+ny1m+n)P(x,y) = (\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}).     * External Division: Replace nn with n-n.     * Harmonic Conjugates: If $P$ divides $AB$ internally and $Q$ divides $AB$ externally in ratio $m:n$, then $P$ and $Q$ are harmonic conjugates. Distances $AP, AB, AQ$ satisfy 2AB=1AP+1AQ\frac{2}{AB} = \frac{1}{AP} + \frac{1}{AQ}.

  • Collinearity of Three Points:     * Conditions: Area of triangle is 0; Slope of $AB$ = Slope of $BC$; Section formula is applicable with a consistent ratio (λ\lambda).

Centers of Triangles

  • Centroid (GG):     * Intersection of medians. Divides each median in ratio 2:12:1 from the vertex.     * Coordinates: (x1+x2+x33,y1+y2+y33)(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}).     * Area Property: $G$ divides a triangle into 3 triangles of equal area and 6 smaller triangles of equal area.

  • Incenter (II):     * Intersection of internal angle bisectors. Center of the circle touching all sides.     * Equidistant from all sides.     * Coordinates: (ax1+bx2+cx3a+b+c,ay1+by2+cy3a+b+c)(\frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c}).

  • Circumcenter (OO):     * Intersection of perpendicular bisectors. Center of circle passing through vertices.     * For a Right Triangle: Midpoint of the hypotenuse.

  • Orthocenter (HH):     * Intersection of altitudes.     * For a Right Triangle: The vertex where the $90^{\circ}$ angle is formed.

  • The Euler Line (OGHO-G-H Rule):     * In any non-equilateral triangle, the Orthocenter ($H$), Centroid ($G$), and Circumcenter ($O$) are collinear.     * Centroid ($G$) divides the segment HOHO in the ratio 2:12:1.     * O-N-G-C mnemonic: Orthocenter ($H$), Nine-point center ($N$), Centroid ($G$), Circumcenter ($O$).

  • Special Triangle Properties:     * Equilateral: All centers coincide (H=G=I=OH = G = I = O).     * Isosceles: All centers are collinear.     * Image Property: The image of the Orthocenter ($H$) with respect to any side lies on the circumcircle.

Slope and Intercepts

  • Slope (mm):     * m=tan(θ)m = \tan(\theta) where θ\theta is the angle with the positive X-axis.     * Two-point form: m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}.

  • Parallel and Perpendicular Lines:     * Parallel: m1=m2m_1 = m_2.     * Perpendicular: m1×m2=1m_1 \times m_2 = -1.

  • Angle Between Two Lines:     * tan(θ)=m1m21+m1m2\tan(\theta) = |\frac{m_1 - m_2}{1 + m_1 m_2}|

  • Line Intercept Properties:     * Equal Intercepts: Slopes m=1m = -1.     * Equal Length of Intercepts: Slope m=±1m = \pm 1.     * Equal Magnitude but Opposite Sign: Slope m=1m = 1.     * Equally Inclined with Axes: Slope m=±1m = \pm 1.

Questions & Discussion

  • Question: How many lines can pass through one point?

  • Response: Infinite lines. If you have both a point and a fixed slope, then exactly one line is defined.

  • Question: What happens to the slope of a vertical line?

  • Response: The angle is 9090^\circ, so tan(90)\tan(90^\circ) is undefined. Thus, the slope is not defined.

  • Question: In a right-angled triangle, where is the circumcenter located?

  • Response: It is exactly at the midpoint of the hypotenuse.