Sequences and Series: Previous Year Questions with Solutions

1.0Introduction

Sequences and Series Previous Year Questions typically cover concepts such as arithmetic progression (AP), geometric progression (GP), harmonic progression (HP), sum of n terms, nth term, and special series like sum of squares and cubes. Common question types include finding specific terms in a sequence, determining the sum of a series, proving identities involving sequences, and solving real-life applications using progression formulas. Solutions often involve the direct application of standard formulas, logical pattern recognition, and algebraic manipulation. Practicing these questions strengthens understanding of patterns in numbers and builds a strong foundation for solving complex mathematical problems efficiently.

2.0Sequences and Series: Previous Year Questions for JEE with Solutions

JEE questions in Sequences and Series often test fundamental concepts involving different types of progressions and their applications. Key areas include arithmetic progression (AP), geometric progression (GP), harmonic progression (HP), and special series like sum of squares, cubes, and telescoping series. Some common types of problems include:

• Arithmetic and Geometric Progressions: Questions on finding the nth term, sum of n terms, insertion of means, and solving word problems using AP/GP formulas.
• Harmonic Progression and Special Series: Problems involving conversion between HP and AP, use of standard summation formulas, and evaluation of complex series.
• Mixed Series and JEE-Type Patterns: Questions that combine concepts of different progressions or involve manipulation of series using algebraic identities.

These questions are designed to test both conceptual clarity and the ability to apply formulas quickly and accurately under time constraints.

Note: In the JEE Main Mathematics exam, you can typically expect 2 to 3 questions from the Sequences and Series chapter.

Related Videos: Sequence and Series for JEE

3.0Key Concepts to Remember – Sequences and Series

1. Types of Sequences:
2. • Arithmetic Progression (AP): Constant difference between consecutive terms.
   • Geometric Progression (GP): Constant ratio between consecutive terms.
   • Harmonic Progression (HP): Reciprocals of terms form an AP.
   • General Sequence: May not follow a fixed pattern; requires pattern recognition.
2. Formulas:

nth term of AP: $a_n=a+(n-1)d$

Sum of first n terms of AP: $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$

nth term of GP: $a_n=a{r}^{n-1}$

Sum of first n terms of GP:

$S_n=a\frac{r^n-1}{r-1},\text{if }r\ne 1$

$S_n=\text{na if r = 1}$

Sum of infinite GP (|r| < 1): $S=\frac{a}{1-r}$

nth term of HP: Take reciprocals of an AP:

$a_n=\frac{1}{a+(n-1)d}$

3. Important Series:

$$\begin{gathered} (\sum _{k=1}^{n}k=\frac{n(n+1)}{2}) \\ (\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}) \\ (\sum _{k=1}^n{k}^3={\left[\frac{n(n+1)}{2}\right]}^2) \end{gathered}$$

4. Useful Techniques:

• Telescoping Series: Break complex series into a form where terms cancel out.
• Mathematical Induction: Useful for proving general formulas or identities.
• Logarithmic Transformation: For simplifying GP or exponential expressions.
• Algebraic Manipulation: Factorization, completing the square, etc., to identify patterns.

5. JEE-Specific Tips:

• Pay close attention to mixed series and questions involving both AP and GP.
• Practice identifying the pattern in non-standard or recursive sequences.
• Always check if the series is finite or infinite, especially in GP problems.

4.0JEE Mains Past Year Questions with Solutions on Sequences and Series

1. Let three real numbers a,b,c be in arithmetic progression and a + 1, b, c + 3 be in geometric progression. If a > 10 and the arithmetic mean of a,b and c is 8, then the cube of the geometric mean of a,b and c is 

(1) 120

(2) 312

(3) 316

(4) 128

Ans. (1)

Sol.

$$\begin{gathered} 2b=a+c,b^2=(a+1)(c+3) \\ \frac{a+b+c}{3}=8\Rightarrow b=8,a+c=16 \\ 64=(a+1)(19-a)=18a-a^2 \\ {a}^2-18a+45=0\Rightarrow (a-15)(a-3)=0(a>10) \\ a=15,c=1,b=8 \\ ((abc{)}^{1/3}{)}^3=abc=120 \end{gathered}$$

2. The value of is $\frac{1\cdot 2^2+2\cdot 3^2+\cdots +100\cdot (101{)}^2}{1^2\cdot 2+2^2\cdot 3+\cdots +100^2\cdot 101}$

$$\begin{gathered} (1)\frac{306}{305} \\ (2)\frac{305}{301} \\ (3)\frac{32}{31} \\ (4)\frac{31}{30} \end{gathered}$$

Ans. (2)

Sol.

$$\begin{gathered} {\sum }_{r=1}^{100}\left(r^3+2r^2+r\right) \\ =\frac{(n(n+1){)}^2}{4}+\frac{2n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2} \\ ={\sum }_{r=1}^{100}(r^3+r^2) \\ ={\left[\frac{n(n+1)}{2}\right]}^2+\frac{n(n+1)(2n+1)}{3}+\frac{3n(n+1)}{6} \\ =\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{2}{3}(2n+1)+1\right] \\ =\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{2(2n+1)}{3}+1\right] \\ =\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{2}{3}(2n+1)+1\right]=\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{2(2n+1)}{3}+1\right] \\ =\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\left(\frac{2}{3}(2n+1)+1\right)\right]=\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{2(2n+1)+3}{3}\right] \\ =\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{4n+5}{3}\right] \\ \frac{100\cdot 101}{2}\left[\frac{100\cdot 101}{2}+\frac{2}{3}\cdot 201+1\right]=\frac{100\cdot 101}{2}\left[\frac{100\cdot 101}{2}+\frac{402}{3}+1\right] \\ =\frac{100\cdot 101}{2}\left[\frac{100\cdot 101}{2}+134+1\right]=\frac{100\cdot 101}{2}\cdot \left(\frac{100\cdot 101}{2}+135\right) \\ =\frac{5185}{5117}=\frac{305}{301} \end{gathered}$$

3. Let the first three terms 2, p and q, with q ≠ 2, of a G.P. be respectively the 7^th, 8^th and 13^th terms of an A.P. If the 5^th term of the G.P. is the n^th term of the A.P., then n is equal to

(1) 151

(2) 169

(3) 177

(4) 163

Ans. (4)

Sol. p² = 2q

2 = a + 6d ...(i)

p = a + 7d ...(ii)

q = a + 12d ...(iii)

p – 2 = d       ((ii) – (i))

q – p = 5d       ((iii) – (ii))

q – p = 5(p – 2)

q = 6p – 10

p² = 2(6p – 10)

p² – 12p + 20 = 0

p = 10, 2

p = 10 ; q = 50

d = 8

a = –46

2, 10, 50, 250, 1250

ar⁴ = a + (n – 1)d

1250 = –46 + (n – 1)8

n = 163

4. If $\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\cdots +\frac{1}{\sqrt{99}+\sqrt{100}}=m\text{and}\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots +\frac{1}{99\cdot 100}=n,$ then the point (m, n) lies on the line

(1) 11(x – 1) – 100(y – 2) = 0 

(2) 11(x – 2) – 100(y – 1) = 0 

(3) 11(x – 1) – 100y = 0 

(4) 11x – 100y = 0 

Ans. (4)

Sol.

$$\begin{gathered} \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\dots +\frac{1}{\sqrt{99}+\sqrt{100}}=m \\ \frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}\dots \frac{\sqrt{99}-\sqrt{100}}{-1}=m \\ \sqrt{100}-1=m\Rightarrow m=9 \\ \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\dots \frac{1}{99\cdot 100}=n \\ \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}\dots \frac{1}{99}-\frac{1}{100}=n \\ 1-\frac{1}{100}=n \\ \frac{99}{100}=n \\ (m,n)=\left(9,\frac{99}{100}\right)\Rightarrow 11(9)-100\left(\frac{99}{100}\right) \\ =99-99=0 \\ \text{ Ans. option (4) }11x-100y=0 \end{gathered}$$

5. Let ABC be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle ABC and the same process is repeated infinitely many times. If P is the sum of perimeters and Q is be the sum of areas of all the triangles formed in this process, then: 

$$\begin{gathered} (1)P^2=36\sqrt{3}Q) \\ (2)P^2=6\sqrt{3}Q \\ (3)P=36\sqrt{3}{Q}^2 \\ (4)P^2=72\sqrt{3} \end{gathered}$$

Ans. (1)

Sol.

$$\begin{gathered} \text{Area of first}\Delta =\frac{\sqrt{3}{\mathrm{a}}^2}{4} \\ \text{Area of second}\Delta =\frac{\sqrt{3}{a}^2}{4}\frac{a^2}{4}=\frac{\sqrt{3}{a}^2}{16} \\ \text{Area of third}\Delta =\frac{\sqrt{3}{\mathrm{a}}^2}{64} \\ \text{sum of area}=\frac{\sqrt{3}{a}^2}{4}\left(1+\frac{1}{4}+\frac{1}{16}\dots \right) \\ Q=\frac{\sqrt{3}{a}^2}{4}\frac{1}{\frac{3}{4}}=\frac{a^2}{\sqrt{3}} \\ perimeterof{1}^{\text{st }}\Delta =3\mathrm{a} \\ perimeterof{2}^{\text{nd }}\Delta =\frac{\text{ 3a }}{2} \\ perimeterof{3}^{\text{rd }}\Delta =\frac{\text{ 3a }}{4} \\ P=3a\left(1+\frac{1}{2}+\frac{1}{4}+\dots \right) \\ P=3a\cdot 2=6a \\ a=\frac{P}{6} \\ Q=\frac{1}{\sqrt{3}}\cdot \frac{P^2}{36} \\ {P}^2=36\sqrt{3}Q \end{gathered}$$

6. If S(x) = (1 + x) + 2(1 + x)² + 3(1 + x)³+.....+ 60(1 + x)⁶⁰, x ≠ 0 and (60)² S(60) = a(b)^b + b, where a, b ∈ N, then (a + b) equal to ______ 

Ans. (3660)

Sol.

$$\begin{gathered} (1+x)S(x)=(1+x{)}^2+2(1+x{)}^3+\cdots +59(1+x{)}^{61}+60(1+x{)}^{61} \\ (1+x)S(x)={\sum }_{k=2}^{61}(k-1)(1+x{)}^k \\ (1+x)S(x)-S(x)=xS(x)={\sum }_{k=1}^{60}k(1+x{)}^{k+1}-{\sum }_{k=1}^{60}k(1+x{)}^k \\ xS(x)=\frac{(1+x)(1+x{)}^{60}-1}{x}-60(1+x{)}^{61} \\ -60S=\frac{61\cdot \left(61^{60}-1\right)}{60}-60\cdot 61^{61} \end{gathered}$$

7. In an increasing geometric progression ol positive terms, the sum of the second and sixth terms is and the product of the third and fifth terms is 49. Then the sum of the 4^th, 6^th and 8^th terms is :-

(1) 96

(2) 78

(3) 91

(4) 84

Ans. (3)

Sol.

$$\begin{gathered} T_2+T_6=\frac{70}{3} \\ ar+a{r}^5=\frac{70}{3} \\ {\mathrm{T}}_3\cdot {\mathrm{T}}_5=49 \\ {\mathrm{ar}}^2.{\mathrm{ar}}^4=49 \\ {a}^2{r}^6=49 \\ a{r}^3=+7,a=\frac{7}{r^3} \\ \mathrm{ar}\left(1+r^4\right)=\frac{70}{3} \\ \frac{7}{r^2}\left(1+r^4\right)=\frac{70}{3},r^2=t \\ \frac{1}{t}\left(1+t^2\right)=\frac{10}{3} \\ 3t^2-10t+3=0 \\ t=3\frac{1}{3} \\ IncreasingG.P.r^2=3,r=\sqrt{3} \\ {T}_4+T_6+T_8 \\ =a{r}^3+a{r}^5+a{r}^7 \\ =a{r}^3\left(1+r^2+r^4\right) \\ =7(1+3+9)=91 \end{gathered}$$

8. Let a, ar, ar², ....... be an infinite G.P. If ${\sum }_{n=0}^{\infty }a{r}^n=57,{\sum }_{n=0}^{\infty }a{r}^{3n}=9747$,  then a + 18r is equal to :

(1) 27

(2) 46

(3) 38

(4) 31

Ans. (4)

Sol.

$$\begin{gathered} {\sum }_{n=0}^{\infty }a{r}^n=57 \\ {\sum }_{n=0}^{\infty }a{r}^n=\frac{a}{1-r},{\sum }_{n=0}^{\infty }a{r}^{3n}=\frac{a}{1-r^3} \\ a+ar+a{r}^2+\cdots =57 \\ \Rightarrow \frac{a}{1-r}=\mathrm{57......}(1) \\ {\sum }_{n=0}^{\infty }a{r}^{3n}=9747 \\ \Rightarrow a+a{r}^3+a{r}^6+\cdots =\frac{a}{1-r^3}=\mathrm{9747........}(II) \\ a{r}^3+a{r}^6+\cdots =9746=\frac{a{r}^3}{1-r^3} \\ {\left(\frac{a}{1-r}\right)}^3=\frac{a}{1-r^3}\cdot a^2{r}^3 \\ {\left(\frac{a}{1-r}\right)}^3=\frac{a^3}{1-r^3}\Rightarrow \frac{57^3}{9717}=a^3\Rightarrow a=19 \\ \frac{a}{1-r}=57,a=19\Rightarrow \frac{19}{1-r}=57\Rightarrow r=\frac{2}{3}\frac{3}{2} \\ a+18r=19+18\cdot \frac{2}{3}=19+12=31 \end{gathered}$$

9. If $\left(\frac{1}{\alpha +1}+\frac{1}{\alpha +2}+\cdots +\frac{1}{\alpha +1012}\right)-\left(\frac{1}{2\cdot 1}+\frac{1}{4\cdot 3}+\frac{1}{6\cdot 5}+\cdots +\frac{1}{2024\cdot 2023}\right)=\frac{1}{2024}$ ,then α is equal to

Ans. (1011)

Sol.  

$$\begin{gathered} {\sum }_{k=1}^{1012}\frac{1}{\alpha +k}=\left(\frac{1}{\alpha +1}+\frac{1}{\alpha +2}+\cdots +\frac{1}{\alpha +1012}\right) \\ {\sum }_{k=1}^{1012}\frac{1}{(2k)(2k-1)}={\sum }_{k=1}^{1012}\left(\frac{1}{2k-1}-\frac{1}{2k}\right) \\ \left(\frac{1}{\alpha +1}+\cdots +\frac{1}{\alpha +1012}\right)-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\cdots +\frac{1}{2023}-\frac{1}{2024}\right)=\frac{1}{2024} \\ \left(\frac{1}{\alpha +1}+\frac{1}{\alpha +2}+\cdots +\frac{1}{\alpha +2012}\right)-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{2023}\right) \\ +\left\{\frac{1}{2024}-2\left(\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2022}\right)\right\}+\frac{1}{2024}+\left(\frac{1}{\alpha +1}+\cdots +\frac{1}{\alpha +1011}\right) \\ =\left(\frac{1}{\alpha +1}+\cdots +\frac{1}{\alpha +2012}\right)-\left(\frac{1}{1}+\frac{1}{2}+\cdots +\frac{1}{2023}\right)=\frac{1}{2024} \\ ，\frac{1}{\alpha +1}+\cdots +\frac{1}{\alpha +2012}=\frac{1}{1012}+\frac{1}{1013}+\cdots +\frac{1}{2023} \\ \Rightarrow \alpha =1011 \end{gathered}$$

10. If the sum of series $\frac{1}{1\cdot (1+d)}+\frac{1}{(1+d)(1+2d)}+\cdots +\frac{1}{(1+9d)(1+10d)}$ is equal to 5, then 50d is equal to :

(1) 20

(2) 5

(3) 15

(4) 10

Ans. (2)

Sol.

$$\begin{gathered} =\frac{1}{d}\left[\frac{(1+d)-1}{1\cdot (1+d)}+\frac{(1+2d)-(1+d)}{(1+d)(1+2d)}+\cdots \right] \\ =\frac{(1+10d)-(1+9d)}{(1+9d)(1+10d)}=\frac{d}{(1+9d)(1+10d)} \\ \frac{1}{d}\left[-\left(\frac{1}{1}-\frac{1}{1+d}\right)+\left(\frac{1}{1+d}-\frac{1}{1+2d}\right)+\cdots \right] \\ \left(\frac{1}{1+9d}-\frac{1}{1+10d}\right)=5 \\ \Rightarrow \frac{1}{d}\left[\frac{1}{1+9d}-\frac{1}{1+10d}\right]=5 \\ \Rightarrow \frac{10d-9d}{(1+9d)(1+10d)}=\frac{d}{(1+9d)(1+10d)}=5d \end{gathered}$$

11. Let 3, a, b, c be in A.P. and 3, a – 1, b + l, c + 9 be in G.P. Then, the arithmetic mean of a, b and c is :

(1) –4

(2) –1

(3) 13

(4) 11

Ans. (4)

Sol.

$$\begin{gathered} 3,a,b,c\to \text{ A.P }\Rightarrow 3,3+d,3+2d,3+3d \\ 3,a-1,b+1,c+9\to \text{ G.P }\Rightarrow 3,2+d,4+2d,12+3d \\ a=3+d(2+d{)}^2=3(4+2d) \\ b=3+2d \\ c=3+3d \\ \text{ If }d=4\text{ G.P }\Rightarrow 3,6,12,24 \\ a=7 \\ b=11 \\ c=15 \\ \frac{a+b+c}{3}=11 \end{gathered}$$

12. The 20^th term from the end of the progression $20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4},\dots ,-129\frac{1}{4}$ is :-

(1) –118

(2) –110

(3) –115

(4) –100

Ans.(3)

Sol.

Sol.

$$\begin{gathered} 20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4},\dots \dots ,-129\frac{1}{4} \\ \text{This is A.P. with common difference} \\ {d}_1=-1+\frac{1}{4}=-\frac{3}{4} \\ -129\frac{1}{4},\dots \dots \dots \dots ,19\frac{1}{4},20 \\ \text{This is also A.P.}\mathrm{a}=-129\frac{1}{4}and\mathrm{d}=\frac{3}{4} \\ \text{Required term =} \\ -129\frac{1}{4}+(20-1)\left(\frac{3}{4}\right) \\ =-129-\frac{1}{4}+15-\frac{3}{4}=-115 \end{gathered}$$

13. If log_e a, log_e b, log_e c  are in an A.P. and log_e a – log_e2b, log_e2b – log_e3c, log_e3c – log_e a are also in an A.P, then a : b : c is equal to  

(1) 9 : 6 : 4

(2) 16 : 4 : 1

(3) 25 : 10 : 4

(4) 6 : 3 : 2

Ans. (1)

Sol.

$$\begin{gathered} {\log }_{e}a,{\log }_{e}b,{\log }_{e}c \\ \therefore b^2=ac.......(i) \\ \log \left(\frac{a}{2b}\right),\log \left(\frac{2b}{3c}\right),\log \left(\frac{3c}{a}\right)\text{are in A.P.} \\ {\left(\frac{2b}{3c}\right)}^2=\frac{a}{2b}\times \frac{3c}{a} \\ \frac{b}{c}=\frac{3}{2} \\ Puttingineq.(i)b^2=a\times \frac{2b}{3} \\ \frac{a}{b}=\frac{3}{2} \end{gathered}$$

14. If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to

(1) 7

(2) 4

(3) 5

(4) 6

Ans. (4)

Sol.

$$\begin{gathered} a+ar+a{r}^2+a{r}^3+\dots +a{r}^{63} \\ =7(a+a{r}^2+a{r}^4+\dots +a{r}^{62}) \\ \Rightarrow \frac{a(1-r^{64})}{1-r}=\frac{7a(1-r^{64})}{1-r^2} \end{gathered}$$

15. Let a and b be be two distinct positive real numbers. Let 11^th term of a GP, whose first term is a and third term is b, is equal to p^th term of another GP, whose first term is a and fifth term is b. Then p is equal to 

(1) 20

(2) 25

(3) 21

(4) 24

Ans. (3)

Sol.

$$\begin{gathered} 1^{\text{st}}\text{ G.P.}\Rightarrow t_1=a,t_5=b=a{r}^4\Rightarrow r^2=\frac{b}{a} \\ {t}_{11}=a{r}^{10}=a(r^2{)}^5=a\cdot {\left(\frac{b}{a}\right)}^5 \\ {2}^{\text{nd}}\text{ G.P.}\Rightarrow T_1=a,T_5=a{r}^4=b \\ {r}^4=\left(\frac{b}{a}\right)\Rightarrow r={\left(\frac{b}{a}\right)}^{1/4} \\ {T}_p=a{r}^{p-1}=a{\left(\frac{b}{a}\right)}^{\frac{p-1}{4}} \\ {t}_{11}=T_p=a{\left(\frac{b}{a}\right)}^5=a{\left(\frac{b}{a}\right)}^{\frac{p-1}{4}} \\ 5=\frac{p-1}{4}\Rightarrow p=21 \end{gathered}$$

16. Let S_n be the sum to n-terms of an arithmetic progression 3, 7, 11, ...... . If $40<\left(\frac{6}{n(n+1)}{\sum }_{k=1}^n{S}_k\right)<42$, then n equals ___ .

Ans. (9)

Sol.

$$\begin{gathered} s_n=3+7+11+\dots \text{n terms} \\ =\frac{n}{2}(6+(n-1)4)=3n+2n^2-2n \\ =2n^2+n \\ {\sum }_{k=1}^n{s}_k={\sum }_{k=1}^n(2k^2+k)=2{\sum }_{k=1}^n{k}^2+{\sum }_{k=1}^{n}k \\ =2\cdot \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2} \\ =n(n+1)\left[\frac{2n+1}{3}+\frac{1}{2}\right] \\ =\frac{n(n+1)(4n+5)}{6} \\ 40<\left(\frac{6}{n(n+1)}{\sum }_{k=1}^n{S}_k\right)<42 \end{gathered}$$

17. Let upto 10 terms and $\alpha =1^2+4^2+8^2+13^2+19^2+26^2+\dots \text{upto 10 terms}\text{and}\beta ={\sum }_{n=1}^{10}{n}^4$. If $4\alpha -\beta =55k+40$, then k is equal to __________.

Ans. (353)

Sol.

$$\begin{gathered} \alpha =1^2+4^2+8^2+\dots \\ Assume:t_n=a{n}^2+bn+c \\ a=\frac{1}{2},b=\frac{3}{2},c=-1 \\ \alpha ={\sum }_{n=1}^{10}{\left(\frac{n}{2}\left(\frac{3n}{2}-1\right)\right)}^2 \\ 4\alpha ={\sum }_{n=1}^{10}(n^2+3n-2{)}^2,\beta ={\sum }_{n=1}^{10}{n}^4 \\ 4\alpha -\beta ={\sum }_{n=1}^{10}(n^4+5n^3-12n+4)=55(353)+40 \end{gathered}$$

18. If the range of f(θ) = $f(\theta )=\frac{{\sin }^4\theta +3{\cos }^2\theta }{{\sin }^4\theta +{\cos }^2\theta }$  is [α, β], then the sum of the infinite G.P., whose first term is 64 and the common ratio is $\frac{\alpha }{\beta }$, is equal to ______.

Ans. (96)

$$\begin{gathered} Sol. \\ f(\theta )=\frac{{\sin }^4\theta +3{\cos }^2\theta }{{\sin }^4\theta +{\cos }^2\theta } \\ f(\theta )=1+\frac{2{\cos }^2\theta }{{\sin }^4\theta +{\cos }^2\theta } \\ f(0)=\frac{2{\cos }^2\theta }{{\cos }^4\theta -{\cos }^2\theta +1} \\ f(\theta )=\frac{2}{{\cos }^2\theta +{\sec }^2\theta -1}+1 \\ f(\theta ){|}_{\min }=1 \\ f(\boxed{}){|}_{\max }=3 \\ S=\frac{64}{1-\frac{1}{3}}=96 \end{gathered}$$