Sequences and Series — Page 15 | JEE Mathematics

Q141

x = \sum\limits_{n = 0}^\infty {{a^n},y = \sum\limits_{n = 0}^\infty {{b^n},z = \sum\limits_{n = 0}^\infty {{c^n}} } }

, where a, b, c are in A.P. and |a|

A x, y, z are in A.P.

B x, y, z are in G.P.

C

{1 \over x}

,

{1 \over y}

,

{1 \over z}

are in A.P.

D

{1 \over x}

+

{1 \over y}

+

{1 \over z}

= 1

-

(a + b + c)

Correct Answer

Option C

Solution

x = \sum\limits_{n = 0}^\infty {{a^n} = {1 \over {1 - a}};\,y = \sum\limits_{n = 0}^\infty {{b^n} = {1 \over {1 - b}};\,z = \sum\limits_{n = 0}^\infty {{c^n} = {1 \over {1 - c}}} } }

Now, a, b, c $\to$ AP 1 $-$ a, 1 $-$ b, 1 $-$ c $\to$ AP

{1 \over {1 - a}},\,{1 \over {1 - b}},\,{1 \over {1 - c}} \to HP

x, y, z $\to$ HP $\therefore$

{1 \over x},{1 \over y},{1 \over z} \to AP

Q142

Let

S_n

denote the sum of first

n

terms of an arithmetic progression. If

S_{20}=790

and

S_{10}=145

, then

\mathrm{S}_{15}-\mathrm{S}_5

is :

A 405

B 390

C 410

D 395

Correct Answer

Option D

Solution

\begin{aligned} &\begin{aligned} & \mathrm{S}_{20}=\frac{20}{2}[2 \mathrm{a}+19 \mathrm{~d}]=790 \\ & 2 \mathrm{a}+19 \mathrm{~d}=79 \quad \text{.... (1)}\\ & \mathrm{~S}_{10}=\frac{10}{2}[2 \mathrm{a}+9 \mathrm{~d}]=145 \\ & 2 \mathrm{a}+9 \mathrm{~d}=29 \quad \text{.... (2)} \end{aligned}\\ &\text{ From (1) and (2) } a=-8, d=5 \end{aligned}

\begin{aligned} & S_{15}-S_5=\frac{15}{2}[2 a+14 d]-\frac{5}{2}[2 a+4 d] \\ & =\frac{15}{2}[-16+70]-\frac{5}{2}[-16+20] \\ & =405-10 \\ & =395 \end{aligned}

Q143

Let

x_1, x_2, x_3, x_4

be in a geometric progression. If

2,7,9,5

are subtracted respectively from

x_1, x_2, x_3, x_4

, then the resulting numbers are in an arithmetic progression. Then the value of

\dfrac{1}{24}\left(x_1 x_2 x_3 x_4\right)

is:

A 18

B 216

C 36

D 72

Correct Answer

Option B

Solution

Given the sequence $x_1, x_2, x_3, x_4$ in geometric progression: $x_1 = a$ $x_2 = ar$ $x_3 = ar^2$ $x_4 = ar^3$ When you subtract 2, 7, 9, and 5 from $x_1, x_2, x_3, x_4$ respectively, the sequence becomes an arithmetic progression.

Thus, the new sequence is: $a - 2$ $ar - 7$ $ar^2 - 9$ $ar^3 - 5$ For these to form an arithmetic progression, the common differences must be equal, so: $(ar - 7) - (a - 2) = (ar^2 - 9) - (ar - 7)$ Simplifying gives: $a(r - 1) - 5 = ar(r - 1) - 2$ $a(r - 1)(r - 1) = -3 \quad (i)$ $(ar - 7) - (a - 2) = (ar^3 - 5) - (ar^2 - 9)$ Simplifying gives: $a(r - 1) - 5 = ar^2(r - 1) + 4$ $a(r - 1)(r^2 - 1) = -9 \quad (ii)$ Using the ratio of equations (ii) and (i): $\dfrac{a(r - 1)(r^2 - 1)}{a(r - 1)(r - 1)} = \dfrac{-9}{-3}$ $r + 1 = 3 \implies r = 2$ Plugging back into equation (i): $a(1)(1) = -3 \implies a = -3$ So, the sequence $x_1, x_2, x_3, x_4$ is: $x_1 = -3$ $x_2 = -6$ $x_3 = -12$ $x_4 = -24$ The expression for $\dfrac{1}{24}(x_1 \cdot x_2 \cdot x_3 \cdot x_4)$ is: $\dfrac{1}{24}((-3) \cdot (-6) \cdot (-12) \cdot (-24)) = 216$

Q144

If

7=5+\dfrac{1}{7}(5+\alpha)+\dfrac{1}{7^2}(5+2 \alpha)+\dfrac{1}{7^3}(5+3 \alpha)+\ldots \ldots \ldots \ldots \infty

, then the value of

\alpha

is :

A

\dfrac{1}{7}

B 1

C

\dfrac{6}{7}

D 6

Correct Answer

Option D

Solution

\begin{aligned} & \text{ Let } S=5+\frac{1}{7}(5+\alpha)+\frac{1}{7^2}(5+2 \alpha)+\ldots \\ & \frac{1}{7} S=\frac{1}{7}(5)+\frac{1}{7^2}(5+\alpha)+\ldots \infty \end{aligned}

\begin{aligned} &\frac{6}{7}(S)=5+\frac{1}{7} \alpha\left(\frac{1}{1-\frac{1}{7}}\right)\\ &6=5+\frac{\alpha}{6} \Rightarrow \alpha=6 \end{aligned}

Q145

The number of terms of an A.P. is even; the sum of all the odd terms is 24 , the sum of all the even terms is 30 and the last term exceeds the first by

\dfrac{21}{2}

. Then the number of terms which are integers in the A.P. is :

A 6

B 4

C 8

D 10

Correct Answer

Option B

Solution

Let the number of terms be $2 n$

\begin{aligned} & n d=6 \\ & (a+(2 n+1) d)-a=\frac{21}{2} \\ & \Rightarrow 2 n d-d=\frac{21}{2} \\ & \Rightarrow 12-\frac{21}{2}=d \end{aligned}

\begin{aligned} & \Rightarrow d=\frac{3}{2} \\ & \therefore n=4 \\ & \therefore \text{ Total terms }=8 \end{aligned}

Q146

Let

x_{1}, x_{2}, \ldots, x_{100}

be in an arithmetic progression, with

x_{1}=2

and their mean equal to 200 . If

y_{i}=i\left(x_{i}-i\right), 1 \leq i \leq 100

, then the mean of

y_{1}, y_{2}, \ldots, y_{100}

is :

A 10051.50

B 10049.50

C 10100

D 10101.50

Correct Answer

Option B

Solution

We have, mean of $x_1, x_2 \ldots \ldots x_{100}=200$ Where, $x_1, x_2 \ldots x_{100}$ are in AP with first term as 2.

\begin{aligned} \text{ Mean } & =200 \\\\ & =\frac{\sum\limits_{i=1}^{100} x_i}{100}=200 \end{aligned}

\begin{aligned} \frac{100}{2} \times[2 \times 2+99 d] =20000 \\\\ \Rightarrow 4+99 d =400 \\\\ \Rightarrow 99 d =396 \\\\ d =4 \end{aligned}

Also,

\begin{aligned} y_i & =i\left(x_i-i\right) \\\\ & =i[2+(i-1) 4-i] \\\\ & =i[3 i-2] \\\\ & =3 i^2-2 i \end{aligned}

\begin{aligned} \text{ Required mean } & =\frac{\sum\limits_{i=1}^{100} y_i}{100} \\\\ & =\frac{1}{100}\left[\sum_{i=1}^{100}\left(3 i^2-2 i\right)\right] \\\\ & =\frac{1}{100}\left[\frac{3 \times 100 \times 101 \times 201}{6}-2 \times \frac{100 \times 101}{2}\right]\\\\ & =\frac{20301}{2}-101 \\\\ & =10049.50 \end{aligned}

Q147

For three positive integers p, q, r,

{x^{p{q^2}}} = {y^{qr}} = {z^{{p^2}r}}

and r = pq + 1 such that 3, 3 log

_yx

, 3 log

_zy

, 7 log

_xz

are in A.P. with common difference

\dfrac{1}{2}

. Then r-p-q is equal to

A 12

B

-

6

C 6

D 2

Correct Answer

Option D

Solution

$x^{p q^{2}}=y^{q r}=z^{p^{2} r}$

3 \log _{y} x=\frac{7}{2}, 3 \log _{z} y=4,7 \log _{x} z=\frac{9}{2}

\begin{aligned} & \Rightarrow x=y^{\frac{7}{6}}, y=z^{\frac{4}{3}}, z=x^{\frac{9}{14}} \\\\ & y^{\frac{7}{6} p q^{2}}=y^{q r}=y^{\frac{3}{4} p^{2} r} \\\\ & \Rightarrow \frac{7}{6} p q^{2}=q r=\frac{3}{4} p^{2} r \\\\ & \therefore 7 p q=6 r, 4 q=3 p^{2} \\\\ & r=p q+1 \\\\ & r=\frac{6 r}{7}+1 \Rightarrow r=7 \\\\ & p q=6 \\\\ & p\left(\frac{3 p^{2}}{4}\right)=6 \\\\ & p=2, q=3 \\\\ & r-p-q=7-5=2 \end{aligned}

Q148

Let

S_{K}=\dfrac{1+2+\ldots+K}{K}

and

\sum\limits_{j=1}^{n} S_{j}^{2}=\dfrac{n}{A}\left(B n^{2}+C n+D\right)

, where

A, B, C, D \in \mathbb{N}

and

A

has least value. Then

A

A+B+C+D

is divisible by 5

B

A+C+D

is not divisible by

B

C

A+B=5(D-C)

D

A+B

is divisible by

\mathrm{D}

Correct Answer

Option D

Solution

\begin{aligned} & \because S_k=\frac{1+2+\ldots+k}{k} \\\\ & =\frac{k(k+1)}{2 k}=\frac{k+1}{2} \end{aligned}

\begin{aligned} & \Rightarrow S_k^2=\left(\frac{k+1}{2}\right)^2=\frac{k^2+1+2 k}{4} \\\\ & \Rightarrow \sum_{j=1}^n S_j^2=\frac{1}{4}\left[\sum_{j=1}^n k^2+\sum_{j=1}^n 1+2 \sum_{j=1}^n k\right] \end{aligned}

\begin{aligned} & =\frac{1}{4}\left[\frac{n(n+1)(2 n+1)}{6}+n+\frac{2 n(n+1)}{2}\right] \\\\ & =\frac{n}{4}\left[\frac{(n+1)(2 n+1)}{6}+1+n+1\right] \\\\ & =\frac{n}{24}\left[2 n^2+3 n+1+6+6 n+6\right] \\\\ & =\frac{n}{24}\left[2 n^2+9 n+13\right] \end{aligned}

On comparing, we get

\mathrm{A}=24, \mathrm{~B}=2, \mathrm{C}=9, \mathrm{D}=13

(A) $A+B+C+D=48$ , which is not divisible by 5.

(B) $\mathrm{A}+\mathrm{C}+\mathrm{D}=46$ , which is divisible by 2(B).

(C) $A+B=26$

\begin{aligned} & 5(\mathrm{D}-\mathrm{C})=5(13-9)=20 \\\\ & \therefore 26 \neq 20 \end{aligned}

(D) $A+B=24+2=26$ , divisible by 13.

Q149

Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second row consists of two balls and so on. If 99 more identical balls are addded to the total number of balls used in forming the equilaterial triangle, then all these balls can be arranged in a square whose each side contains exactly 2 balls less than the number of balls each side of the triangle contains. Then the number of balls used to form the equilateral triangle is :-

A 262

B 190

C 157

D 225

Correct Answer

Option B

Solution

As triangle is equilateral so each side of triangle has n-balls.

At the top of the triangle there is 1 ball then next line has 2 balls.

Similarly last line has n balls.

$\therefore$ Total no of balls used to create this triangle = 1 + 2 + 3 + ..............+ n =

{{n\left( {n + 1} \right)} \over 2}

Here Number of balls in each side of square is = (n – 2) $\therefore$ Total no of balls used to create this square = (n – 2)2 According to the question,

{{n\left( {n + 1} \right)} \over 2}

+ 99 = (n – 2)2 $\Rightarrow$ n2 - 9n - 190 = 0 $\Rightarrow$ (n - 19) (n + 10) = 0 $\therefore$ n = 19 So total balls used to form triangle =

{{n\left( {n + 1} \right)} \over 2}

=

{{19\left( {20} \right)} \over 2}

= 190

Q150

If

\operatorname{gcd}~(\mathrm{m}, \mathrm{n})=1

and

1^{2}-2^{2}+3^{2}-4^{2}+\ldots . .+(2021)^{2}-(2022)^{2}+(2023)^{2}=1012 ~m^{2} n

then

m^{2}-n^{2}

is equal to :

A 220

B 200

C 240

D 180

Correct Answer

Option C

Solution

Given $\operatorname{gcd}(m, n)=1$ and

\begin{aligned} & \Rightarrow 1^2-2^2+3^2-4^2+\ldots .+(2021)^2-(2022)^2+(2023)^2 \\ & =1012 m^2 n \\\\ & \Rightarrow 1^2-2^2+3^2-4^2+\ldots .+(2021)^2-(2022)^2+(2023)^2 \\ & =1012 m^2 n \\\\ & \Rightarrow(1-2)(1+2)+(3-4)(3+4)+\ldots .(2021-2022) \\ & (2021+2022)+(2023)^2=(1012) m^2 n \\\\ & \Rightarrow(-1)(1+2)+(-1)(3+4)+\ldots .+ \\ & \quad(-1)(2021+2022)+\left(2023^2\right)=(1012) m^2 n \end{aligned}

$\begin{aligned} & \Rightarrow(-1)[1+2+3+4+\ldots+2022]+(2023)^2 =1012 m^2 n \\\\ & \Rightarrow(-1)\left[\dfrac{(2022) \cdot(2022+1)}{2}\right]+(2023)^2=(1012) m^2 n \\\\ & \Rightarrow(-1)\left[\dfrac{(2022)(2023)}{2}\right]+(2023)^2=(1012) m^2 n \\\\ & \Rightarrow(2023)[2023-1011]=(1012) m^2 n \\\\ & \Rightarrow m^2 n=2023 \\\\\end{aligned}$ $\begin{array}{lc}\Rightarrow m^2 n=289 \times 7 \\\\ \Rightarrow m^2=289 \text{ and } n=7 \\\\ \Rightarrow m=17 \text{ and } n=7(\text{ such that } \operatorname{gcd}(m, n)=1) \\\\ \therefore m^2-n^2=289-49=240\end{array}$