Sequences and Series — Page 7 | JEE Mathematics

Q61

If each term of a geometric progression

a_1, a_2, a_3, \ldots

with

a_1=\dfrac{1}{8}

and

a_2 \neq a_1

, is the arithmetic mean of the next two terms and

S_n=a_1+a_2+\ldots . .+a_n

, then

S_{20}-S_{18}

is equal to

A

-2^{15}

B

2^{15}

C

-2^{18}

D

2^{18}

Correct Answer

Option A

Solution

Let

r^{\prime}

th term of the GP be

a^{n-1}

. Given,

\begin{aligned} & 2 a_r=a_{r+1}+a_{r+2} \\ & 2 a r^{n-1}=a r^n+a r^{n+1} \\ & \frac{2}{r}=1+r \\ & r^2+r-2=0 \end{aligned}

Hence, we get,

r=-2

(as

r \neq 1

) So,

\mathrm{S}_{20}-\mathrm{S}_{18}=

(Sum upto 20 terms) $-$ (Sum upto 18 terms)

=\mathrm{T}_{19}+\mathrm{T}_{20}

\mathrm{~T}_{19}+\mathrm{T}_{20}=\mathrm{ar}^{18}(1+\mathrm{r})

Putting the values

\mathrm{a}=\frac{1}{8}

and

\mathrm{r}=-2

; we get

T_{19}+T_{20}=-2^{15}

Q62

Let

\mathrm{a}_{\mathrm{n}}

be the

\mathrm{n}^{\text{th }}

term of the series

5+8+14+23+35+50+\ldots

and

\mathrm{S}_{\mathrm{n}}=\sum\limits_{k=1}^{n} a_{k}

. Then

\mathrm{S}_{30}-a_{40}

is equal to :

A 11280

B 11290

C 11310

D 11260

Correct Answer

Option B

Solution

Let $\mathrm{S}_n=5+8+14+23+\ldots .+a_n$ and $\mathrm{S}_n=0+5+8+14+\ldots .+a_n$ On subtracting, we get

\begin{aligned} & 0=5+3+6 \ldots-a_n \\\\ & \Rightarrow a_n=5+3+6+9+\ldots(n-1) \text{ terms } \\\\ & =5+\left[\frac{(n-1)}{2}(6+(n-2) 3)\right] \end{aligned}

\begin{aligned} & =5+\left[\frac{(n-1)}{2}(6+3 n-6)\right] \\\\ & =5+\frac{(n-1)(3 n)}{2} \\\\ & =\frac{10+3 n^2-3 n}{2} \end{aligned}

\begin{aligned} & \text{ So, } a_{40}=\frac{3(40)^2-3(40)+10}{2} \\\\ & =\frac{4800-120+10}{2}=2345 \end{aligned}

\begin{aligned} & \text{ Now, } S_n=\sum_{k=1}^n a_k \\\\ & \Rightarrow S_{30}=\frac{3 \sum_{n=1}^{30} n^2-3 \sum_{n=1}^{30} n+10 \sum_{n=1}^{30} 1}{2} \\\\ & =\frac{3 \times(30)(30+1)(60+1)}{12}-\frac{3 \times 30 \times 31}{4} +\frac{10 \times 30}{2} \end{aligned}

\begin{aligned} & =\frac{28365-1395+300}{2}=\frac{27270}{2} \\\\ & =13635 \\\\ & \therefore S_{30}-a_{40}=13635-2345=11290 \end{aligned}

Q63

Let

2^{\text{nd }}, 8^{\text{th }}

and

44^{\text{th }}

terms of a non-constant A. P. be respectively the

1^{\text{st }}, 2^{\text{nd }}

and

3^{\text{rd }}

terms of a G. P. If the first term of the A. P. is 1, then the sum of its first 20 terms is equal to -

A 990

B 980

C 960

D 970

Correct Answer

Option D

Solution

\begin{aligned} & 1+d, \quad 1+7 d, 1+43 d \text{ are in GP } \\ & (1+7 d)^2=(1+d)(1+43 d) \\ & 1+49 d^2+14 d=1+44 d+43 d^2 \\ & 6 d^2-30 d=0 \\ & d=5 \\ & S_{20}=\frac{20}{2}[2 \times 1+(20-1) \times 5] \\ & \quad=10[2+95] \\ & \quad=970 \end{aligned}

Q64

The sum of the series

\dfrac{1}{1-3 \cdot 1^2+1^4}+\dfrac{2}{1-3 \cdot 2^2+2^4}+\dfrac{3}{1-3 \cdot 3^2+3^4}+\ldots

up to 10 -terms is

A

\dfrac{45}{109}

B

-\dfrac{55}{109}

C

\dfrac{55}{109}

D

-\dfrac{45}{109}

Correct Answer

Option B

Solution

General term of the sequence,

\begin{aligned} & \mathrm{T}_{\mathrm{r}}=\frac{\mathrm{r}}{1-3 \mathrm{r}^2+\mathrm{r}^4} \\ & \mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\mathrm{r}^4-2 \mathrm{r}^2+1-\mathrm{r}^2} \\ & \mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\left(\mathrm{r}^2-1\right)^2-\mathrm{r}^2} \\ & \mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\left(\mathrm{r}^2-\mathrm{r}-1\right)\left(\mathrm{r}^2+\mathrm{r}-1\right)} \\ & \mathrm{T}_{\mathrm{r}}=\frac{\frac{1}{2}\left[\left(\mathrm{r}^2+\mathrm{r}-1\right)-\left(\mathrm{r}^2-\mathrm{r}-1\right)\right]}{\left(\mathrm{r}^2-\mathrm{r}-1\right)\left(\mathrm{r}^2+\mathrm{r}-1\right)} \\ & =\frac{1}{2}\left[\frac{1}{\mathrm{r}^2-\mathrm{r}-1}-\frac{1}{\mathrm{r}^2+\mathrm{r}-1}\right] \end{aligned}

Sum of 10 terms,

\sum\limits_{\mathrm{r}=1}^{10} \mathrm{~T}_{\mathrm{r}}=\frac{1}{2}\left[\frac{1}{-1}-\frac{1}{109}\right]=\frac{-55}{109}

Q65

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

A 7

B 6

C 5

D 4

Correct Answer

Option B

Solution

\begin{aligned} & a+a r+a r^2+a r^3+\ldots .+a r^{63} \\ & =7\left(a+a r^2+a r^4 \ldots .+a r^{62}\right) \\ & \Rightarrow \frac{a\left(1-r^{64}\right)}{1-r}=\frac{7 a\left(1-r^{64}\right)}{1-r^2} \\ & r=6 \end{aligned}

Q66

In an A.P., the sixth term

a_6=2

. If the product

a_1 a_4 a_5

is the greatest, then the common difference of the A.P. is equal to

A

\dfrac{2}{3}

B

\dfrac{5}{8}

C

\dfrac{3}{2}

D

\dfrac{8}{5}

Correct Answer

Option D

Solution

\begin{aligned} & a_6=2 \Rightarrow a+5 d=2 \\ & a_1 a_4 a_5=a(a+3 d)(a+4 d) \\ & =(2-5 d)(2-2 d)(2-d) \\ & f(d)=8-32 d+34 d^2-20 d+30 d^2-10 d^3 \\ & f^{\prime}(d)=-2(5 d-8)(3 d-2) \end{aligned}

\mathrm{d}=\frac{8}{5}

Q67

The value of

\dfrac{1 \times 2^2+2 \times 3^2+\ldots+100 \times(101)^2}{1^2 \times 2+2^2 \times 3+\ldots .+100^2 \times 101}

is

A

\dfrac{305}{301}

B

\dfrac{306}{305}

C

\dfrac{32}{31}

D

\dfrac{31}{30}

Correct Answer

Option A

Solution

\begin{aligned} & \frac{1 \times 2^2+2 \times 3^2+\ldots+100 \times(101)^2}{1^2 \times 2+2^2 \times 3+\ldots+100^2 \times 101} \\ & \Rightarrow \frac{\sum\limits_{n=1}^{100} n(n+1)^2}{\sum\limits_{n=1}^{100} n^2(n+1)} \end{aligned}

\begin{aligned} & \Rightarrow \frac{\sum\limits_{n=1}^{100} n^3+2 n^2+n}{\sum\limits_{n=1}^{100} n^3+n^2} \\ & =\frac{\left(\frac{100(101)}{2}\right)^2+\frac{2 \cdot 100(101)(201)}{6}+\frac{100(101)}{2}}{\left(\frac{100(101)}{2}\right)^2+\frac{100(101)(201)}{6}} \\ & =\frac{300(101)+4(201)+6}{300(101)+2(201)}=\frac{5185}{5117}=\frac{305}{301} \end{aligned}

Q68

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

A 120

B 316

C 312

D 128

Correct Answer

Option A

Solution

\begin{aligned} & 2 b=a+c \quad \text{.... (1)}\\ & b^2=(a+1)(c+3) \quad \text{.... (2)}\\ & \frac{a+b+c}{3}=8 \quad \text{.... (3)} \end{aligned}

\begin{aligned} \Rightarrow & \frac{3 b}{3}=8 \\ & b=8 \\ \Rightarrow \quad & a c+3 a+c+3=64 \end{aligned}

\begin{aligned} & 3 a+c+a c=61 \quad \text{... (4)}\\ & a+c=16 \\ & c=16-a \end{aligned}

from equation (4)

\begin{aligned} & 3 a+16-a+a(16-a)=61 \\ & \Rightarrow \quad(a-15)(a-3)=0 \\ & \quad a=15(a>10) \\ & \Rightarrow \quad a=15, b=8, c=1 \\ & \left((a \cdot b \cdot c)^{\frac{1}{3}}\right)^3=15 \times 8 \times 1=120 \end{aligned}

Q69

Let the first three terms 2, p and q, with

q \neq 2

, of a G.P. be respectively the

7^{\text{th }}, 8^{\text{th }}

and

13^{\text{th }}

terms of an A.P. If the

5^{\text{th }}

term of the G.P. is the

n^{\text{th }}

term of the A.P., then

n

is equal to:

A 151

B 177

C 163

D 169

Correct Answer

Option C

Solution

\begin{aligned} & \text{ Let } p=2 r, q=2 r^2 \\ & T_7=2, T_8=2 r, T_{13}=2 r^2 \\ & d=2 r-2=2(r-1) \\ & 2 r^2=T_7+6 d=2+6(2)(r-1)=12 r-10 \\ & \Rightarrow r^2-6 r+5=0 \\ & \Rightarrow(r-1)(r-5)=0 \\ & \therefore r=1,5 \\ & r=1 \text{ (rejected) as } q \neq 2 \\ & \therefore r=5 \end{aligned}

5^{\text{th }}

term of G.P

=2 . r^4=2.5^4

Let

1^{\text{st }}

term of A.P

b a=a, d=8

2=a+(6)(8) \Rightarrow a=-46

\mathrm{n}^{\text{th }}

term of A.P

=-46+(n-1) 8=8 n-54

\begin{aligned} & 2.5^4=8 n-54 \\ & \Rightarrow 1250+54=8 n \\ & \Rightarrow n=\frac{1304}{8}=163 \end{aligned}

Q70

Let

A B C

be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle

A B C

and the same process is repeated infinitely many times. If

\mathrm{P}

is the sum of perimeters and

Q

is be the sum of areas of all the triangles formed in this process, then :

A

\mathrm{P}^2=72 \sqrt{3} \mathrm{Q}

B

\mathrm{P}^2=36 \sqrt{3} \mathrm{Q}

C

\mathrm{P}=36 \sqrt{3} \mathrm{Q}^2

D

\mathrm{P}^2=6 \sqrt{3} \mathrm{Q}

Correct Answer

Option B

Solution

\triangle A B C

is an equilateral triangle having side

=a

unit Now, perimeter $=$ sum of all sides

=3 a

Area

=\frac{\sqrt{3}}{4} a^2

Now,

\ln \triangle D E F, D E=\frac{a}{2}=E F=D F

\begin{aligned} & \text{ Perimeter }=3 \times \frac{a}{2}=\frac{3 a}{2} \\ & \text{ Area }=\frac{\sqrt{3}}{4} \times\left(\frac{a}{2}\right)^2=\frac{\sqrt{3} a^2}{16} \end{aligned}

Now,

P=3 a+\frac{3 a}{2}+\frac{3 a}{4}+\cdots

Q=\frac{\sqrt{3}}{4} a^2+\frac{\sqrt{3}}{16} a^2+\frac{\sqrt{3}}{64} a^2+\cdots

P=\frac{3 a}{1-\frac{1}{2}}=3 a \times 2 \Rightarrow P=6 a \quad \text{.... (i)}

Q=\frac{\frac{\sqrt{3}}{4} a^2}{1-\frac{1}{4}}=\frac{4}{3} \times \frac{\sqrt{3}}{4} a^2 \quad Q=\frac{\sqrt{3}}{3} a^2 \quad \text{.... (ii)}

From equation (i) & (ii)

\begin{aligned} & P=6 a \\ & Q=\frac{\sqrt{3}}{3} a^2 \\ & Q=\frac{\sqrt{3}}{3} \times \frac{P^2}{36} \\ & P^2=36 \sqrt{3} Q \end{aligned}