### Complete the a.p series with given information - class-XI

 Description: complete the a.p series with given information Number of Questions: 75 Created by: Tags: sequences and series progressions sequence, progression and series binomial theorem, sequence and series maths arithmetic progressions
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$a$ should be $5$ to make the series $2,a,8,...$ to be in AP

1. True

2. False

Correct Option: A
Explanation:

Given series $2,a,8$

The condition to be in AP is
$2b=a+c\2a=2+8\2a=10\a=5$

If the solution as $\cos p\theta +\cos q\theta=0$ are in $AP$ then the common difference is

1. $\dfrac {\pi}{p+q}$ or $\dfrac {\pi}{p-q}$

2. $\dfrac {\pi}{p+q}$ or $\dfrac {\pi}{2(p-q)}$

3. $\dfrac {\pi}{2(p+q)}$ or $\dfrac {\pi}{2(p-q)}$

4. $None\ of\ these$

Correct Option: A

The $n^{th}$ term of the series $3+7+14+24+.....$ is

1. $\dfrac{n-1}{2}(3n+2)$

2. $3+\dfrac{n-1}{2}(3n+2)$

3. $1+\dfrac{n-1}{2}(3n+2)$

4. $3+\dfrac{(n+1)(3n+2)}{2}$

Correct Option: B

A line passes through the variable point $A(\lambda +1,2\lambda)$ meets the lines $7x+y-16=0,\ 5x-y-8=0,\ x-5y+8=0$ at $b,c,d$ respectively. Then $AC, AB, AD$ are in

1. A.P

2. G.P

3. H.P

4. None of these

Correct Option: A

If $T _n$ denotes the nth terms of the series.$2+3+6+11+18+..........$ , then $T _50$ is :

1. $49^2-1$

2. $49^2$

3. $50^2+1$

4. $49^2+2$

Correct Option: D
Explanation:

$2+3+6+11+18+.................$

$\begin{matrix} { T _{ n } }=a{ x^{ 2 } }+bx+c \ { T _{ 1 } }=a+b+c=2 \ { T _{ 2 } }=a+b+c=3 \ { T _{ 3 } }=a+b+c=6 \ 3a+b=1 \ 5a+b=3 \ 2a=2 \ a=1 \ b=-2 \ c=3 \ \end{matrix}$
${T _n} = {n^2} - 2n + 3$
${T _{50}} = {\left( {50} \right)^2} - 2\left( {50} \right) + 3$
${T _{50}} = {\left( {49} \right)^2} + 2$
Hence,
Option $D$ is correct answer.

If the m+n, n+p, p+n terms of an AP are a, b, c respectively, then m(b-c)+n(c-a)+p(a-b) is

1. 1

2. a+b+c

3. m+n+p

4. 0

Correct Option: D
Explanation:
$T _{m}=A+(m-1)d=a$
$T _{n}=A+(n-1)d=b$
$T _{p}=A+(p-1)d=c$
From these equations, we get,
$a-b=(m-n)d$
$b-c=(n-p)d$
and $c-a=(p-m)d$
Now,
$m(b-c)+n(c-a)+p(a-b)=m(n-p)d+n(p-m)d+p(m-n)d$
$=d(mn-mp+np-nm+pm-pn)$
$=d \times 0=0$

Let $a,b,c$ be in AP and $k\neq 0$ be a real number. WHich of trhe following are correct ?
1.$ka,kb,kc$ are in Ap
2. $k-a,k-b,k-c$ are in AP
3. $\dfrac{a}{k},\dfrac{b}{k},\dfrac{c}{k}$ are in AP
Select the correct answer using the code given below :

1. 1 and 2 only

2. 2 and 3 only

3. 1 and 3 only

4. 1, 2 and 3 only

Correct Option: D
Explanation:

$a, b, c \in A.P. \quad \left( \text{Given} \right)$

Therefore,
$(1) \quad\text{when a constant term is multiplied to each term of an A.P. then the resultant is also an A.P.}$
$ka, kb, kc \in A.P., \quad k \ne 0$

$(2)\quad\text{when a constant is subtracted from each term of an A.P. then the resultant is also an A.P.}$
$\Rightarrow k - a, k-b, k - c \in A.P.$

$(3)\quad\text{when a constant is divided from each term of an A.P. then the resultant is also an A.P.}$
$\Rightarrow \cfrac{a}{k}, \cfrac{b}{k}, \cfrac{c}{k} \in A.P.$
Hence all statements are correct.

Find the next term of the series:
$22, 26, 29, 31,$ ..........

1. $32$

2. $33$

3. $34$

4. $35$

Correct Option: A
Explanation:

Lets find the difference between the numbers of the series:

$22$       $26$       $29$       $31$       $x$.....
$4$          $3$         $2$        $1$
Since, the difference is in A.P.
$\therefore x=31+1=32$
Hence, the answer is $32$.

$\displaystyle a^{2}\left ( b+c \right ),b^{2}\left ( c+a \right ),c^{2}\left ( a+b \right )$ provided $\displaystyle \sum ab\neq 0.$ if $a=b=c$ then above series is in:

1. AP

2. GP

3. HP

4. AGP

Correct Option: A,B,C,D

If the $m^{th}$ term and the $n^th$ term of an AP are respectively $\displaystyle \frac { 1 }{ n }$ and $\displaystyle \frac { 1 }{ m }$, then the $mn^{th}$ term of the AP is

1. $\displaystyle \frac { 1 }{ mn }$

2. $\displaystyle \frac { m }{ n }$

3. $\displaystyle 1$

4. $\displaystyle \frac { n }{ m }$

Correct Option: C
Explanation:

Let a and d be the first term and common difference of an AP.
Since, $\displaystyle { T } _{ m }=\frac { 1 }{ n }$
$\displaystyle \therefore a+\left( m-1 \right) d=\frac { 1 }{ n }$....(i)
and $\displaystyle { T } _{ n }=\frac { 1 }{ m }$
$\displaystyle \Rightarrow a+\left( n-1 \right) d=\frac { 1 }{ m }$.....(ii)
On solving Eqs. (i) and (ii), we get
$\displaystyle a=\frac { 1 }{ mn } and\quad d=\frac { 1 }{ mn }$
$\displaystyle \therefore \quad { T } _{ mn }=a+\left( mn-1 \right) d$

$\displaystyle =\frac { 1 }{ mn } +\frac { \left( mn-1 \right) }{ mn }$

$\displaystyle =\frac { mn }{ mn } =1$

Which one is an example of A.P. property?

1. Constant $a$ is added to each term of an A.P. will form a new A.P with different common difference

2. Constant $a$ is subtracted to each term of an A.P. will form a new A.P with different common difference

3. Constant $a$ is divided to each term of an A.P. will not form a new A.P with same common difference

4. Constant $a$ is added to each term of an A.P. will form a new A.P with same common difference

Correct Option: D
Explanation:

If you will subtract or add a constant number to each terms of an AP, then later sequence will also be in AP with same common difference

Hence option 'D' is correct choice

Identify the property of A.P. used in the sequence:

$(3 - x), (5 - x), (7 - x), (9 - x)$

1. $3$ is a constant subtracted from the sequence

2. $-x$ is a constant subtracted from the sequence

3. $x$ is a constant subtracted from the sequence

4. Number is a constant subtracted from the sequence

Correct Option: C
Explanation:
Given sequence is $(3-x),(5-x),(7-x),(9-x)$
Here $3,5,7,9$ are in AP with common difference $2$
Using the property of AP, if we subtract the same no. from each term, then it will remain in AP.
i.e. $3-x , 5-x, 7-x, 9-x$ is all equal to $2$ which will be in AP.

In an arithmetic progression the sum of two terms equidistant from the beginning and the end is always _____ to the sum of the first and last terms.

1. equal

2. unequal

3. different

4. various

Correct Option: A
Explanation:

Let $a$ be first term $d$ be common difference of an AP having $n$ number of terms.

So $m$th term from the beginning is, $a _m=a+(m-1)d$
and $m$ the term from end is $a _{n-m}= a+(n-1)d-(m-1)d=a+(n-m)d$
So $a _m+a _{n-m}=2a+(n-1)d=[a]+[a+(n-1)d]=$ sum of first and last term

$\dfrac{1}{x},\dfrac{2}{x},\dfrac{3}{x},....$ is a property of

1. G.P.

2. A.P.

3. A.S.

4. H.P.

Correct Option: B
Explanation:

$\dfrac{1}{x},\dfrac{2}{x},\dfrac{3}{x},....$ is a property of A.P.
Here the constant x is divided from each term of an A.P. with same common difference.

In which property the sum of two terms equidistant from the beginning and the end is always same or equal to the sum of the first and last terms?

1. A.P.

2. G.P.

3. H.P.

4. AGP

Correct Option: A
Explanation:

Let $a$ be first term $d$ be common difference of an AP having $n$ number of terms.

So $m$th term from the beginning is, $a _m=a+(m-1)d$
and $m$ the term from end is $a _{n-m}= a+(n-1)d-(m-1)d=a+(n-m)d$
So $a _m+a _{n-m}=2a+(n-1)d=[a]+[a+(n-1)d]=$ sum of first and last term

Hence option 'A' is correct choice

How many natural numbers are there between $23$ and $100$ which are exactly divisible by $24$?

1. $8$

2. $11$

3. $12$

4. $13$

5. None of these

Correct Option: D
Explanation:

Required numbers are $24,30,36,40,.....96$
This is an A.P. in which $a=24,d=6,l=96$
Let the number of terms in it be $n$.
Then ${t} _{n}=96$ $\Rightarrow$ $a+(n-1)d=96$
$\Rightarrow$ $24+(n-1)\times 6=96$
$\Rightarrow$ $(n-1)=12$
$\Rightarrow$ $n=13$
Required number of numbers $=13$

The sum of first $10$ terms and $20$ terms of an AP are $120$ and $440$ respectively. What is the first term?

1. $2$

2. $3$

3. $4$

4. $5$

Correct Option: B
Explanation:

Let the first term be $a$ and common difference be $d$.
So, sum of first $10$ terms $=\dfrac { 10 }{ 2 } (2a+(10-1)d)$

$\implies 120 =5(2a+9d)$
$\implies 24=2a+9d$ .............. $(i)$
Sum of first 20 terms $=\frac { 20 }{ 2 } (2a+(20-1)d)$
$\implies 440 =10(2a+19d)$
$\implies 44=2a+19d$ ......... $(ii)$
Subtracting equation (i) from (ii) gives
$20=10d$
$\implies d=2$
Common difference =2
Substituting in $(i)$, we get
$a=3$
Hence, option B is correct

$T _m$ denotes the number of Triangles that can be formed with the vertices of a regular polygon of $m$ sides.If $T _m+ _1-T _m=15$ , then $m$

1. 3

2. 6

3. 9

4. 12

Correct Option: B
Explanation:

$T _m+ _1-T _m=15$

$\Rightarrow ^{(m+1)}C _3-^{(m)}C _3=15$

By verification $m=6$ by solve

Which term of A.P. $20, 19\displaystyle\frac{1}{4}, 18\frac{1}{2}$,..... is first negative term?

1. $!8$th

2. $15$th

3. $28$th

4. $27$th

Correct Option: C
Explanation:

$20, \displaystyle 19\frac{1}{4}, 18\frac{1}{2}, ....$
or $\displaystyle 20, \frac{77}{4}, \frac{37}{2}, ....$
$a=20$
$d=\displaystyle\frac{77}{4}-20=\frac{-3}{4}$
Let $n^{th}$ term of A.P. be first negative term
So, $20+(n-1)\left(\displaystyle\frac{-3}{4}\right)<0$
$\Rightarrow 80-3n+3<0$
$\Rightarrow 3n>83$
$\Rightarrow n > 27\displaystyle\frac{2}{3}$
Hence, $28^{th}$ term is first negative term.
(Option $3$).

${ T } _{ m }$ denotes the number of triangles that can be formed with the vertices of a regular polygon of m sides. If ${ { T } _{ m+1 } }-{ { T } _{ m } }=15,$ then $m=$

1. $3$

2. $6$

3. $9$

4. $12$

Correct Option: B

If $a,b,c$ are distinct and the roots of $\left( b-c \right) { x }^{ 2 }+\left( c-a \right) x+\left( a-b \right) =0$ are equal, then $a,b,c$ are in

1. Arithmetic progression

2. Geometric progression

3. Harmonic progression

4. Arithmetico-Geometric progression

Correct Option: A
Explanation:

Clearly $x=1$ is a solution
$\therefore$  product of the roots $=\dfrac { a-b }{ b-c }$
$\therefore \left( 1 \right) \left( 1 \right) =\dfrac { a-b }{ b-c }$
$\Longrightarrow b-c=a-b$
$\Longrightarrow2b=a+c\Longrightarrow a,b,c$ are in Arithmetic progression.

Say true or false.
In an $A.P$., sum of terms equidistant from the beginning and end is constant and is equal to the sum of the first and last term.

1. True

2. False

Correct Option: A
Explanation:

Consider mth term of an AP
$t _m=a+(m-1)d$
Now, consider (n-m)th term:
$t _{n-m}=a+(n-m-1)d$
Sum $= 2a+(n-1)d$

$= a +a+(n-1)d$
$= a+l$
Therefore, it is true that Sum of the terms is equal to the sum of the first and last terms.

If $a,b,c$ are in $A.P.$, then the straight lines $ax+by+c=0$ wil always pass through the point ..........

1. $(1,2)$

2. $(1,5)$

3. $(3,2)$

4. None of these

Correct Option: A
Explanation:

$A,B, C$ in $AP$

$ax + by + c = 0$ __(I)

$\therefore 2b = a + c$

$a - 2b + c = 0$ __(II)

comparing (I) and (II),

$x = 1, y = -2$

$\therefore (1, -2)$ is the fixed point

An AP consists of $15$ terms. The three middle most terms is $69$ and the last three terms is $123$. Find the A.P.

1. 2,5,8,11.....44

2. 2,4,8,....62

3. 2,3,4....16

4. none

Correct Option: A
Explanation:

The $three$ middle most terms of $15$ terms is $69.$

${a} _{7}+{a} _{8}+{a} _{9}=69$
${a} _{13}+{a} _{14}+{a} _{15}=123$
$a+6d+a+7d+a+8d=69\Longrightarrow 3a+21d=69$
$a+12d+a+13d+a+14d=123\Longrightarrow 3a+39d=123$

$3a+21d=69$
$3a+39d=123$

$18d=54,\d=3,\a=2.$
$2,5,8,11,\dots 44.$

If $x,y,z$ are $p^{th},q^{th}$ and $r^{th}$ terms respectively of a $G.P$., then $x^{q-r}\cdot y^{r-p}\cdot z^{p-q}$ is simplified to

1. $1$

2. $0$

3. $xyz$

4. $None\ of\ these$

Correct Option: A
Explanation:

$x = A{R^{p - 1}}$

$y = A{R^{q - 1}}$
$z = A{R^{r - 1}}$
${x^{q - r}}{y^{r - p}}{z^{p - q}}$
$= {\left( {A{R^{p - 1}}} \right)^{q - r}}{\left( {A{R^{q - 1}}} \right)^{r - p}}{\left( {A{R^{r - 1}}} \right)^{p - q}}$
$= {A^{\left( {q - r + r - p + p - q} \right)}}{R^{\left[ {\left( {p - 1} \right)\left( {q - r} \right) + \left( {q - 1} \right)\left( {r - p} \right) + \left( {r - 1} \right)\left( {p - q} \right)} \right]}}$
$= {A^0}{R^{\left[ {pq - pr - q + r + qr - pq - r + p + pr - qr - p + q} \right]}}$
$= {A^0}{R^0}$
$=1$

If $f _n(x)=\frac{sinx}{cos3x}+\frac{sin3x}{cos3^2x}+\frac{sin3^2x}{cos3^3x}+..........+ \frac{sin3^{n-1}x}{cos3^nx}$ then $f _2$

1. $0$

2. $1$

3. $-1$

4. $3$

Correct Option: C

The largest term to common to the sequences $1,11,21,31,... to 100$ terms and $31,36,41,46, ...to 100$ terms is

1. $511$

2. $471$

3. $281$

4. None of these

Correct Option: A
Explanation:
The first sequence is $1+11+21+31+41+51+61+71+81+91+...............$ upto $100th$ terms
last term will be =$1+(100-1)\times 10=991$
and for the second sequence  $31+36+41+46+51+56+.................$ upto $100th$ term
last term will be=$31+(100-1)\times 5=526$
Now,
we have the common terms as $31+51+71$ and $n$ term of this sequence wil be =$31+(n-1)\times 20$
$=>20n+11$
and this will be less than $526$
So,
$=>20n+11<526$
$=>20n<526-11$
$=>n<\dfrac{515}{20}=25.75$
so $n=25$
Hence the largest common term is =$20\times 25\times 11=511$

If $a^{2},b^{2},c^{2}$ are in $AP$, then which of the following are in $AP$?

1. $(b+c),(c+a),(a+b)$

2. $\dfrac{1}{b+c},\dfrac{1}{c+a},\dfrac{1}{a+b}$

3. $\dfrac{b+c}{a},\dfrac{c+a}{b},\dfrac{a+b}{c}$

4. $\dfrac{1}{a^{2}},\dfrac{1}{a^{2}},\dfrac{1}{c^{2}}$

Correct Option: B
Explanation:
$2b^2=a^2+c^2$
$b^2+b^2=a^2+c^2$
$b^2-a^2=c^2-b^2$
$(b-a)(b+a)=(c-b)(c+b)$
$\dfrac { (b-a) }{ (c+b) } =\dfrac { (c-b) }{ (b+a) }$
Dividing both sides by $(c + a)$

$\dfrac { (b-a) }{ (c+b)(c+a) } =\dfrac { (c-b) }{ { (b+a)(c+a) } }$

$\dfrac { (b+c)-(c+a) }{ (b+c)(c+a) } =\dfrac { (c+a)-(a+b) }{ (a+b)(c+a) }$

$\dfrac { 1 }{ (c+a) } -\dfrac { 1 }{ b+c } =\dfrac { 1 }{ (a+b) } -\dfrac { 1 }{ c+a }$

$\dfrac { 2 }{ (c+a) } =\dfrac { 1 }{ (a+b) } -\dfrac { 1 }{ (b+c) }$

Therefore $\dfrac { 1 }{ (a+b) } ,\dfrac { 1 }{ (b+c) } ,\dfrac { 1 }{ (c+a) }$ are in AP.

If $a,b$ and $c$ are in $A.P.,$ then $\dfrac{(a-c)^{2}}{b^{2}-ac}=$ ?

1. $1$

2. $2$

3. $3$

4. $4$

Correct Option: B
Explanation:

$a,b,c$ are in A.P

So $2b=a+c$
Then
$\cfrac { { \left( a-c \right) }^{ 2 } }{ { b }^{ 2 }-ac }$
$=\cfrac { { \left( a-c \right) }^{ 2 } }{ { b }^{ 2 }-ac }$
$\cfrac { { a }^{ 2 }+{ c }^{ 2 }-2ac }{ { \left( \cfrac { a+c }{ 2 } \right) }^{ 2 }-ac } =\cfrac { { a }^{ 2 }+{ c }^{ 2 }-2ac }{ \left( \cfrac { { a }^{ 2 }+{ c }^{ 2 }-2ac }{ 2 } \right) } =2$

If $a, b, c$ are in $A.P.$ then $\left|\begin{matrix} x+1 & x+2 & x+a \ x+2 & x+3 & x+b \ x+3 & x+4 & x+c \end{matrix}\right|$

1. $1$

2. $0$

3. $-1$

4. $2$

Correct Option: B
Explanation:
$a, b, c in A.P.$
$b=a+d$
$c=a+2d$ where $d$is the common difference
$=\begin{vmatrix} x+1 & x+2 & x+d \\ x+2 & x+3 & x+a+d \\ x+3 & x+4 & x+a+2d \end{vmatrix}$
$=\begin{vmatrix} 1 & 2 & a \\ 2 & 3 & a+d \\ 3 & 4 & a+2d \end{vmatrix}+\begin{vmatrix} x & x & x \\ x & x & x \\ x & x & x \end{vmatrix}\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \downarrow \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0$
$=\begin{vmatrix} 1 & 2 & a \\ 2 & 3 & a+d \\ 3 & 4 & a+2d \end{vmatrix}$
$1\begin{vmatrix} 3 & a+d \\ 4 & a+2d \end{vmatrix}-2\begin{vmatrix} 2 & a+d \\ 3 & a+2d \end{vmatrix}+a\begin{vmatrix} 2 & 3 \\ 3 & 4 \end{vmatrix}$
$=3a+6d-4a-4d-4a-8d+6a+6d-a$
$=(3a-4a-4a+6a-a)+(6d-4d-3d+6d)$
$=0+0$
$=0$
$B$ is correct

If $x \in R,$ the numbers ${2^{1 + x}} + {2^{1 - x}},b/2,{36^x} + {36^{ - x}}$ form an A.P. , then $b$ may lie in the interval

1. $\left[ {16,\infty } \right)$

2. $\left[ {6,\infty } \right)$

3. $\left[ {\infty , - 6} \right)$

4. $\left[ {6,12} \right)$

Correct Option: B
Explanation:

Given

$2^{1+x}+2^{1-x}, \dfrac b2 ,36^x+36^{-x}$ form an AP

The condition to be in AP is

$2\times \left(\dfrac b2\right)=2^{1+x}+2^{1-x}+36^x+36^{-x}$

$b=2.2^x+2.\dfrac1{2^x}+36^x+\dfrac 1{36^x}$

$b=2\left(2^x+\dfrac 1{2^x}\right)+\left( 36^x+\dfrac 1{36^x}\right)$

Let $2^x=y \quad 36^x=k$

$b=2\left(y+\dfrac 1y\right)+\left(k+\dfrac 1k\right)$

The min value of $f(x)+\dfrac 1{f(x)}$ is $2$

The max value is $\infty$

$\implies 2(2)+2 \leq b\leq 2(\infty)+(\infty)$

$\implies 6\leq b\leq\infty$

$\implies b\in [6,\infty)$

State the whether given statement is true or false
For a positive integer n,let $S(n)=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+.....+\dfrac{1}{2^n-1}$. Then prove that $S(100)<100$.

1. True

2. False

Correct Option: A
Explanation:

We have,

$S(n)=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+.....+\dfrac{1}{2^n-1}$

$S(n)=1+\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}\right)+\left(\dfrac{1}{8}+\dfrac{1}{9}+.......\right)+.....+\dfrac{1}{2^n-1}$

$S(n)=1+\left(\dfrac{1}{2}+\dfrac{1}{2^2-1}\right)+\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{2^3-1}\right)+\left(\dfrac{1}{8}+\dfrac{1}{9}+.......+\dfrac{1}{2^4-1}\right)+.....+\dfrac{1}{2^n-1}$

Since,
$\left(\dfrac{1}{2}+\dfrac{1}{3}\right)<1$

$\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}\right)<1$

$\dfrac{1}{2^n-1}<1$

So,
$S(n)=1+1+1+......+1\ n(times)$
$S(n)<n$

Therefore,
$S(100)<100$

If $9^{th}$ term of an A.P. be zero then the ratio of its $2022^{th}$ and $10^{th}$ term is.......

1. $2013:1$

2. $1:2013$

3. $2013:8$

4. $8:2013$

Correct Option: A
Explanation:
General term of $A.P\ \Rightarrow \ at (n-1)d$
$\therefore \ a+(9-1)d=0$
$9+8d=0$
$a=-8d$
$\therefore \ \dfrac {a+(2022-1)d}{a+(10-1)d}$
$\Rightarrow \ \dfrac {-8d+2021d}{-8d+9d}$
$=\dfrac {2013d}{d}=\boxed {2013:1}$
$2013:1$, Option $A$ is correct

If the angles  $A,B,C$ of a $\triangle ABC$ are in $A.P.$, then:-

1. ${c}^{2}={a}^{2}+{b}^{2}-ab$

2. ${b}^{2}={a}^{2}+{c}^{2}-ac$

3. ${c}^{2}={a}^{2}+{b}^{2}$

4. None of these

Correct Option: B

If a, b, c are in A.P., then  $a ^ { 3 } + c ^ { 3 } - 8 b ^ { 3 }$ is equal to:

1. $2 a b c$

2. -$6 a b c$

3. $4 a b c$

4. none of these

Correct Option: B
Explanation:
$a,b,c$ are in A.P
$\Rightarrow\,b-a=c-b$
$\Rightarrow\,2b=a+c$
${a}^{3}+{c}^{3}-8{b}^{3}$
$={a}^{3}+{c}^{3}-{\left(2b\right)}^{3}$
$={a}^{3}+{c}^{3}-{\left(a+c\right)}^{3}$
$={a}^{3}+{c}^{3}-{a}^{3}-{c}^{3}-3ac\left(a+c\right)$
$=-3ac\left(2b\right)=-6abc$

If $\dfrac{1}{b-c},\dfrac{1}{c-a},\dfrac{1}{a-b}$ be consecutive terms of an AP then $(b-c)^2,(c-a)^2,(a-b)^2$ will be in ?

1. GP

2. AP

3. HP

4. None of these

Correct Option: B
Explanation:

if $a,b,c$ is in $AP$ then we know that,

$b-a=c-b=d$ where $d$ is the common difference
A/Q,$\dfrac { 1 }{ c-a } -\dfrac { 1 }{ b-c } =\dfrac { 1 }{ a-b } -\dfrac { 1 }{ c-a } \quad \left( \dfrac { 1 }{ b-c } ,\dfrac { 1 }{ c-a } ,\dfrac { 1 }{ a-b } \quad are\quad in\quad AP \right) \ \Rightarrow \dfrac { b-c-c+a }{ (c-a)(b-c) } =\dfrac { c-a-a-b }{ (a-b)(c-a) } \ \Rightarrow { a }^{ 2 }-2ac-{ b }^{ 2 }+2bc={ b }^{ 2 }-2ab-{ c }^{ 2 }+2ac\ \Rightarrow { a }^{ 2 }-2ac+{ c }^{ 2 }-{ c }^{ 2 }-{ b }^{ 2 }+2bc={ b }^{ 2 }-2ab+{ a }^{ 2 }-{ a }^{ 2 }-{ c }^{ 2 }+2ac\ \Rightarrow \left( { c }^{ 2 }-2ac+a^{ 2 } \right) -\left( { b }^{ 2 }-2bc+{ c }^{ 2 } \right) =\left( { a }^{ 2 }-2ab+{ b }^{ 2 } \right) -\left( { c }^{ 2 }-2ac+{ a }^{ 2 } \right) \ \Rightarrow { \left( c-a \right) }^{ 2 }-{ \left( b-c \right) }^{ 2 }={ \left( a-b \right) }^{ 2 }-{ \left( c-a \right) }^{ 2 }\$
$\therefore { \left( b-c \right) }^{ 2 },{ \left( c-a \right) }^{ 2 },{ \left( a-b \right) }^{ 2 }\quad are\quad in\quad AP$

If we divide $20$ into four parts which are in A.P  such that product of the first and the fourth is to the product of the second and the third is the same as $2$:$3$ then the smallest part is

1. $1$

2. $2$

3. $3$

4. $4$

Correct Option: B
Explanation:

Let the four parts be $a-3d,\,a-d,\,a+d$ and $a+3d$

Hence, $(a-3d)+(a-d)+(a+d)+(a+3d)=20$
$\Rightarrow$  $4a=20$
$\therefore$  $a=5$

It is also given that,
$(a-3d)(a+3d):(a-d)(a+d)=2:3$

$\Rightarrow$  $(a^2-9d^2):(a^2-d^2)=2:3$

$\Rightarrow$  $\dfrac{a^2-9d^2}{a^2-d^2}=\dfrac{2}{3}$

$\Rightarrow$  $3(a^2-9d^2)=2(a^2-d^2)$
$\Rightarrow$  $3a^2-27d^2=2a^2-2d^2$
$\Rightarrow$  $3a^2-2a^2=27d^2-2d^2$
$\Rightarrow$  $a^2=25d^2$
$\Rightarrow$  $(5)^2=25d^2$                        [ Substituting value of $a$ ]
$\Rightarrow$  $25=25d^2$
$\Rightarrow$  $d^2=1$
$\therefore$  $d=\pm 1$

Case $I:$ When $d=1$
$\Rightarrow$  $a-3d=5-3=2$
$\Rightarrow$  $a-d=5-1=4$
$\Rightarrow$  $a+d=5+1=6$
$\Rightarrow$  $a+3d=5+3=8$

$\therefore$  The four numbers are $2,4,6$ and $8$

Case $II:$ When $d=-1$
$\Rightarrow$  $a-3d=5+3=8$
$\Rightarrow$  $a-d=5+1=6$
$\Rightarrow$  $a+d=-5-1=4$
$\Rightarrow$  $a+3d=5-3=2$

$\therefore$  The four numbers are $8,6,4$ and $2$

$\therefore$  In both cases we can see the smallest value is $2$

The mean of a data set consisting of $20$ observations is $40$. If one observation $53$ was wrongly recorded as $33$, then the correct mean will be:

1. $41$

2. $49$

3. $40.5$

4. $42.5$

Correct Option: A
Explanation:

$Mean=\dfrac{S}{n}$

where S=sum of all observations
n=number of observations

hence, $40=\dfrac{S}{20}\Rightarrow S=800$

but Since, 53 is recorded as 33 so we need to add $(53-33=20)$ to get the correct mean which is
$Mean _{correct}=\dfrac{800+20}{20}=41$

If $log2,log({ 2 }^{ x }-1)and\quad log({ 2 }^{ x }+3)$ are in A.P., then x is equal to :

1. $\dfrac { 5 }{ 2 }$

2. ${ log } _{ 2 }5$

3. ${ log } _{ 3 }2$

4. ${ log } _{ 5 }2$

Correct Option: B
Explanation:

$\log { 2 } ,\log { \left( { 2 }^{ x }-1 \right) } ,\log { \left( { 2 }^{ x }-3 \right) }$ are in AP

$\log { \left( { 2 }^{ x }-1 \right) } =\log { 2 } +\log { \left( { 2 }^{ x }-3 \right) }$
${ \log { \left( { 2 }^{ x }-1 \right) } }^{ 2 }=\log { \left[ 2.\left( { 2 }^{ x }+3 \right) \right] }$
$\quad { \left( { 2 }^{ x }-1 \right) }^{ 2 }={ 2 }^{ x+1 }+6$
${ \left( { 2 }^{ x } \right) }^{ 2 }+1-2.{ 2 }^{ x }=2.{ 2 }^{ x }+6$
${ \left( { 2 }^{ x } \right) }^{ 2 }-4.{ 2 }^{ x }-5=0\Rightarrow { \left( { 2 }^{ x } \right) }^{ 2 }-5.{ 2 }^{ x }+{ 2 }^{ x }-5=0\Rightarrow { 2 }^{ x }({ 2 }^{ x }-5)+1({ 2 }^{ x }-5)=0\Rightarrow ({ 2 }^{ x }-5)({ 2 }^{ x }+1)=0$
${ 2 }^{ x }+1\neq 0,{ 2 }^{ x }+5=0\Rightarrow { 2 }^{ x }=5$
taking log of base 2
$x\log _{ 2 }{ 2 } =\log _{ 2 }{ 5 }$
$x=\log _{ 2 }{ 5 }$

If $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P., then $(\frac{1}{a}+\frac{1}{b}-\frac{1}{c})(\frac{1}{b}+\frac{1}{c}-\frac{1}{a})$ is equal to:

1. $\frac{4}{ac}-\frac{3}{b^{2}}$

2. $\frac{b^{2}-ac}{a^{2}b^{2}c^{2}}$

3. $\frac{4}{ac}-\frac{1}{b^{2}}$

4. none of these

Correct Option: A

Suppose in $\Delta ABC$, ex-radii $r _1,r _2,r _3$ are H.P. then the sides $a,b,c$

1. will be in A.P.

2. will be in H.P.

3. will be in G.P.

4. $a+2b=c$

Correct Option: A

If the ratio of sum of n terms of two sequences is (3n+8):(7n+15), then the ratio of their $12^{th}$ term is ---------.

1. 16 : 7

2. 7 : 16

3. 74 : 169

4. None of the above

Correct Option: A
Explanation:

Let $a,A$ be first term and $d,D$ be common difference and $S _n,s _n$ be there respective sum.

$\cfrac{s _n}{S _n}=\cfrac{(n/2)[2a+(n-1)d]}{(N/2)[2A+(N-1)D]}=\cfrac{3n+8}{7n+15}$
Put $n=23$, we get
$\cfrac{a+11d}{A+11D}=\cfrac{77}{176}=\cfrac{t _{12}}{T _{12}}$

Let ${a _1},{a _2},{a _3}.....$ and ${b _1},{b _2},{b _3}......$ be AP such that ${a _1}=25,{b _1}=75$ and ${a _{100}} + {b _{100}} = 100$. Then,

1. The difference between successive terms in progression $a$ is opposite the difference in progression $b$

2. ${a _n} + {b _n} = 100$ for any n

3. $({a _1} + {b _1}),({a _2} + {b _2}),({a _3} + {b _3}),....$ are in AP

4. $\sum\limits _{r = 1}^{100} {({a _r} + {b _r})} = 10000$

Correct Option: C

If the roots of palynomial $P ( x ) = x ^ { 3 } - 3 x ^ { 2 } + k x + 4$ are in $A P ,$ then $\left| k \right|$. Has the value equal to

1. 0

2. 1

3. 2

4. 3

Correct Option: C
Explanation:

Given roots of the polynomial are in AP

let the roots of the polynomial be $a-d,a,a+d$
$\quad a-d+a+a+d=-\frac { -3 }{ 1 } \ \Rightarrow 3a=3\ \Rightarrow a=1$
so, $a=1$ is one of the roots of the equation
$\quad p\left( 1 \right) ={ 1 }^{ 3 }-3\times { 1 }^{ 2 }+k+4=0\ \Rightarrow k+2=0\ \Rightarrow k=-2$
$\left| k \right| =2$

If the non-zero terms $x , y, z$ are in $AP$ and $\tan ^ { -1 } x, \tan ^ { - 1 } y, \tan ^ { - 1 } x$ are also $AP$ then

1. $x = y = z$

2. $n ^ { 2 } y = z$

3. $z ^ { 2 } = x y$

4. $y ^ { 2 } = \frac { 1 } { x 2 }$

Correct Option: A
Explanation:

Given $x,yz$ are in $AP$

So, $y=x=2-y$
$2y=x+2 \quad -(1)$
Similarly,
$\tan ^{ -1 }{ x } ,\tan ^{ -1 }{ y }$ and $\tan ^{ -1 }{ z }$ are also in $AP$.
So, $2\tan ^{ -1 }{ y } =\tan ^{ -1 }{ x } +\tan ^{ -1 }{ z }$
$\Rightarrow \tan ^{ -1 }{ (\cfrac { 2y }{ 1-{ y }^{ 2 } } ) } =\tan ^{ -1 }{ (x+\cfrac { z }{ 1-xz } ) } \ \Rightarrow \cfrac { 2y }{ 1-{ y }^{ 2 } } =x+\cfrac { z }{ 1-xz } \ \Rightarrow x+\cfrac { z }{ 1-{ y }^{ 2 } } =x+\cfrac { z }{ 1-xz } \ \Rightarrow \cfrac { 1 }{ 1-{ y }^{ 2 } } =\cfrac { 1 }{ 1-xz } \ \Rightarrow 1-xz=1-{ y }^{ 2 }\ \Rightarrow 1-xz=1-{ { \cfrac { x+z }{ 2 } } }^{ 2 }\ \Rightarrow { x }^{ 2 }+{ z }^{ 2 }+2xz=4xz\ \Rightarrow { x }^{ 2 }+{ z }^{ 2 }+2xz-4xz=0\ \Rightarrow { x }^{ 2 }+{ z }^{ 2 }-2xz=0\ \Rightarrow { (x-z) }^{ 2 }=0$
$\Rightarrow x=z$ put in $(1)$
$\Rightarrow 2y=x+2\ \Rightarrow 2y=2x\ \Rightarrow y=x\ x=y=z$

If roots of the equation $(a-b)x^{2}+(c-a)x+(b-c)=0, a \neq b \neq c$ are equal, then $a,b,c$ are in

1. $A.P$

2. $H.P$

3. $G.P$

4. $None\ of\ these$

Correct Option: D
Explanation:

We have,

Given equation is

$\left( a-b \right){{x}^{2}}+\left( c-a \right)x+\left( b-c \right)=0$

On comparing that,

$A{{x}^{2}}+Bx+C=0$

Now,

$A=\left( a-b \right)$

$B=\left( c-a \right)$

$C=\left( b-c \right)$

Roots are equal

Then,

$D=0$

${{B}^{2}}-4AC=0$

$\Rightarrow {{\left( c-a \right)}^{2}}-4\left( a-b \right)\left( b-c \right)=0$

$\Rightarrow {{c}^{2}}+{{a}^{2}}-2ac=4\left( ab-ac-{{b}^{2}}+bc \right)$

$\Rightarrow {{c}^{2}}+{{a}^{2}}-2ac=4ab-4ac-4{{b}^{2}}+4bc$

$\Rightarrow {{c}^{2}}+{{a}^{2}}-2ac+4ac=4ab-4{{b}^{2}}+4bc$

$\Rightarrow {{\left( c+a \right)}^{2}}=4b\left( a-b+c \right)$

If a, b, c are in AP then $a+\frac{1}{bc}$, $b+\frac{1}{ca}$, $c+\frac{1}{ab}$ are in

1. AP

2. GP

3. HP

4. none of these

Correct Option: A
Explanation:

$a,b,c$ are in AP
$\Rightarrow (abc+1)a, (abc+1)b,(abc+1)c$ are in AP
$\Rightarrow \dfrac{(abc+1)a}{abc},\dfrac{(abc+1)b}{abc},\dfrac{(abc+1)c}{abc}$ are in AP
$\Rightarrow a+\dfrac{1}{bc},b+\dfrac{1}{ac},c+\dfrac{1}{ab}$ are in AP
$\therefore$ Ans. is option A.

Let $a _1, a _2,....a _{10}$ be in AP, and $h _1, h _2,...., h _{10}$ be in HP. If $a _1=h _1=2$ and $a _{10}=h _{10}=3$, then $a _4h _7$ is?

1. $2$

2. $3$

3. $5$

4. $6$

Correct Option: D
Explanation:

${ a } _{ 10 }={ a } _{ 1 }+9{ d } _{ 1 }\qquad \Longrightarrow { d } _{ 1 }=\cfrac { 3-2 }{ 9 } =\cfrac { 1 }{ 9 }$

Now, ${ a } _{ 4 }={ a } _{ 1 }+3{ d } _{ 1 }=2+3\times \left( \cfrac { 1 }{ 9 } \right) =\cfrac { 7 }{ 3 } \ \therefore \left( \cfrac { 1 }{ { h } _{ 10 } } \right) =\left( \cfrac { 1 }{ { h } _{ 1 } } \right) +9{ d } _{ 2 }\ \Longrightarrow \left( \cfrac { 1 }{ 3 } \right) =\left( \cfrac { 1 }{ 2 } \right) +9{ d } _{ 2 }\qquad \Longrightarrow { d } _{ 2 }=\left( \cfrac { -1 }{ 54 } \right)$
Now, $\left( \cfrac { 1 }{ { h } _{ 7 } } \right) =\left( \cfrac { 1 }{ { h } _{ 1 } } \right) +6{ d } _{ 2 }=\cfrac { 1 }{ 2 } +6\left( \cfrac { -1 }{ 54 } \right) \ \left( \cfrac { 1 }{ { h } _{ 7 } } \right) =\cfrac { 7 }{ 18 }$
So, $a _4h _7=\cfrac{7}{3}\times \cfrac{18}{7}=6$

Which term of the sequence $72, 70, 68, 66, ...$ is $40$ ?

1. 12

2. 15

3. 17

4. 19

Correct Option: C
Explanation:

The given sequence is an A.P. with first term $a=72$ and common difference $d=-2$. Let its nth term be 40.

$a _n=40$
$a+(n-1)d=40$
$72+(n-1)(-2)=40$
$72-2n+2=40$
$2n=34$
$n=17$
Hence, 17th term of the sequence is 40.

If $1,\,{\log _y}x,\,{\log _z}y,\, - \,15{\log _{x}z}$ are in $A.P.$ , then

1. ${z^3} = x$

2. $x = {y^{ - 1}}$

3. ${z^{ - 3}} = y$

4. $x = {y^{ - 1}} = {z^3}$

5. All the above

Correct Option: E
Explanation:

Let $d$ be the common difference of the $A.P.$

Then,
$\log _yx=1+d$
$\Rightarrow$  $x=y^{1+d}$                     ----- ( 1 )

$\log _zy=1+2d$
$\Rightarrow$  $y=z^{1+2d}$                   ------ ( 2 )

$-15\log _xz=1+3d$
$\Rightarrow$  $z=x^{\frac{-(1+3d)}{15}}$             ------ ( 3 )

$x=y^{1+d}=z^{(1+2d)(1+d)}=x^{\tfrac{-(1+d)(1+2d)(1+3d)}{15}}$

$\Rightarrow$  $(1+d)(1+2d)(1+3d)=-15$

$\Rightarrow$  $6d^3+11d^2+6d+16=0$

$\Rightarrow$  $(d+2)(6d^2-d+8)=0$

$\Rightarrow$  $d=-2$

Substituting value of $d$ we get,

$\Rightarrow$  $x=y^{-1}=z^3=x^{\tfrac{1}{3}}$ or

$\Rightarrow$  $x=y^{-1}=z^3,\,y=z^{-3}$

The series $4,13,22,31,40,......$ is in

1. $A.P$

2. $G.P$

3. $H.P$

4. $None\ of\ these$

Correct Option: A
Explanation:

The given series is $4,13,22,31,40,....$

The difference between 2 consecutive numbers is $13-4=9\\22-13=9$
So the difference is constant throughout the series
So the series is in $AP$

a proper option (a), (b), (c) or (d) from given options and write in the box given that so that the statement becomes correct : (All the problems refer to A.P)
${ T } _{ 3 }=8,{ T } _{ 7 }=24,$ then ${ T }$

1. -4

2. 28

3. 32

4. 36

Correct Option: A

Select the correct option.
The first term of an $AP$ is $p$ and the common difference is $q$, then its $10^{th}$ term is

1. $q + 9p$

2. $p - 9q$

3. $p + 9q$

4. $2p + 9q$

Correct Option: C
Explanation:

First term of $AP = P$

Common difference of $AP= q$

$10^{th}$ term of $AP = p + (10 - 1) q$

$= p + 9q$

Select the correct option.
The value of $x$ for which $2x, (x + 10)$ and $(3x + 2)$ aree the three consecutive terms of an AP, is

1. $6$

2. $-6$

3. $18$

4. $-18$

Correct Option: C
Explanation:

Consecutive terms of $AP \Rightarrow 2x, (x + 10 ) , (3x + 2)$

According to the properties

$2(x + 10) = 2x + 3x + 2$

$4x + 20 = 5x + 2$

$x = 18$

$\displaystyle \frac{b+c-a}{a}, \frac{c+a-b}{b}, \frac{a+b-c}{c}$ are in A.P., then $\displaystyle \frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in

1. A.P.

2. H.P

3. G.P

4. A.G.P

Correct Option: A
Explanation:

Given,

$\dfrac{b+c-a}{a},\dfrac{c+a-b}{b},\dfrac{a+b-c}{c}$ one in A.P

Now,

$\because \dfrac{b+c-a}{a},\dfrac{c+a-b}{b},\dfrac{a+b-c}{c}$ are in A.P

$\therefore \dfrac{b+c-a}{a}+2,\dfrac{c+a-b}{b}+2,\dfrac{a+b-c}{c}+2$, must be  in A.P

$\therefore \dfrac{b+c-a+2a}{a},\dfrac{c+a-b+2b}{b},\dfrac{a+b-c+2c}{c}$ are in A.P

$\therefore \dfrac{a+b+c}{a},\dfrac{a+b+c}{b},\dfrac{a+b+c}{c}$ are in A.P

$\because \dfrac{a+b+c}{a},\dfrac{a+b+c}{b},\dfrac{a+b+c}{c}$ are in A.P

$\therefore \dfrac{1}{(a+b+c)}\times \dfrac{(a+b+c)}{a},\dfrac{1}{(a+b+c)}\times \dfrac{(a+b+c)}{b},\dfrac{1}{(a+b+c)}\times \dfrac{(a+b+c)}{c}$ are in A.P

$\therefore \dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ are in A.P

Let $S _n$ be the sum of all integers k such that $2^n < k < 2^{n-1}$, for n > 1, Then $9$ divides $S _n$ if and only if

1. $n$ is odd

2. $n$ is of the form $3k+1$

3. $n$ is even

4. $n$ is of the form $3k +2$

Correct Option: C
Explanation:

Number of integers between 2$^n$ and $2^{2n+ 1} - 2^n - 11$
and I term $= 2^n + 1$
last term $= 2^{n + 1} - 1$
$\therefore \displaystyle S _n = \frac{(2^{n + 1} - 2^n - 1)}{2} [2^n + 1 + 2^{n + 1} - 1]$
$\displaystyle \frac{(2^{n + 1} - 2^n - 1)}{2} (2^n)(1 + 2)$
$= (2^n - 1) \displaystyle \frac{(2^n).3}{2}$
$S _n= 9 \lambda; \lambda \varepsilon I$
$\therefore 3 \times 2^{n - 1} \times (2^n - 1) = 9 \lambda$
$2^{n - 1} \times (2^n - 1) = 3 \lambda$
$2^n (2^n - 1) = 6 \lambda$
It is possible when n is even.

The sum of the three numbers in A.P is $21$ and the product of the first and third number of the sequence is $45$. What are the three numbers?

1. $5, 7$ and $9$

2. $9, 7$, and $5$

3. $3, 7$, and $11$

4. Both (1) and (2)

5. None of these

Correct Option: D
Explanation:

Let the numbers are be $a - d, a, a + d$
Then $a - d + a + a + d = 21$
$3a = 21$
$a = 7$
and $(a - d)(a + d) = 45$
$a^2 - d^2 = 45$
$d^2 = 4$
$d=\pm 2$
Hence, the numbers are $5, 7$ and $9$ when $d = 2$ and $9, 7$ and $5$ when $d = -2$. In both the cases numbers are the same.

The sum of $10$ numbers is $100$. The first term is $1$. Find its common difference.

1. $2$

2. $1$

3. $3$

4. $4$

Correct Option: A
Explanation:
The sum of first $n$ terms of arithmetic series formula can be written as,
$S _n = \dfrac{n}{2} [2a + (n - 1)d]$ ............ (1)
$n =$ number of terms $= 10$
$S _n = 100$
First term, $a = 1$
Common difference, $d = ?$
From $(1)$, we have
$100 = \dfrac{10}{2} [2 \times 1 + (10 - 1)d]$
$100 = 5[2 + 9d]$
$100 = 10 + 45d$
$100 - 10 = 45d$
$90 = 45d$
$d = \dfrac{90}{45} = 2$
The common difference is $2$.

The sum of first $10$ terms and $20$ terms of an AP are $120$ and $440$ respectively. What is the common difference?

1. $1$

2. $2$

3. $3$

4. $4$

Correct Option: B
Explanation:

Let the first term be $a$ and common difference be $d$.
So, sum of first $10$ terms $=\dfrac { 10 }{ 2 } (2a+(10-1)d)$

$\implies 120 =5(2a+9d)$
$\implies 24=2a+9d$ .............. $(i)$
Sum of first 20 terms $=\frac { 20 }{ 2 } (2a+(20-1)d)$
$\implies 440 =10(2a+19d)$
$\implies 44=2a+19d$ ......... $(ii)$
Subtracting equation (i) from (ii) gives
$20=10d$
$\implies d=2$
Common difference =2
Hence, option B is correct.

If ${ a } _{ 1 },{ a } _{ 2 },{ a } _{ 3 },\dots$ are terms of AP such that ${ a } _{ 1 }+{ a } _{ 5 }+{ a } _{ 10 }+{ a } _{ 15 }+{ a } _{ 20 }+{ a } _{ 24 }=225$, then the sum of first $24$ terms is

1. $9\times { 10 }^{ 2 }$

2. $9\times { 10 }^{ 3 }$

3. $10\times { 9 }^{ 2 }$

4. $10\times { 9 }^{ 3 }$

Correct Option: A
Explanation:

We know that the sum of terms of AP equidistant from the beginning and end is always same and it is always equal to the sum of first and last terms.
$\Rightarrow { a } _{ 1 }+{ a } _{ 24 }={ a } _{ 6 }+{ a } _{ 20 }={ a } _{ 10 }+{ a } _{ 15 }$
$\because { a } _{ 1 }+{ a } _{ 5 }+{ a } _{ 10 }+{ a } _{ 15 }+{ a } _{ 20 }+{ a } _{ 24 }=225$
$\therefore 3\left( { a } _{ 1 }+{ a } _{ 24 } \right) =225\Rightarrow { a } _{ 1 }+{ a } _{ 24 }=75$
$\therefore { S } _{ 24 }=\dfrac { 24 }{ 2 } \left( { a } _{ 1 }+{ a } _{ 24 } \right)$   $\left[ \because { S } _{ n }=\dfrac { n }{ 2 } \left( { a } _{ 1 }+{ a } _{ n } \right) \right]$
$=12\left( 75 \right) =900=9\times { 10 }^{ 2 }$

$x _{1}, x _{2}, x _{3}, ....$ are in A.P.
If $x _{1} + x _{7} + x _{10} = -6$ and $x _{3} + x _{8} + x _{12} = -11$, then $x _{3} + x _{8} + x _{22} = ?$

1. $-21$

2. $-15$

3. $-18$

4. $-31$

Correct Option: A
Explanation:

Let the first term be $a$ and common difference be $d$

$x _{1}+x _{7}+x _{10}=-6$
$\Rightarrow (a)+(a+6d)+(a+9d)=-6$
$\Rightarrow 3a+15d=-6 \rightarrow \text{eqn}(1)$

$x _{3}+x _{8}+x _{12}=-11$
$\Rightarrow (a+2d)+(a+7d)+(a+11d)=-11$
$\Rightarrow 3a+20d=-11 \rightarrow \text{eqn}(2)$
Solving eqn (1) and (2), we get $a=3$ and $d=-1$

Now, $x _{3}+x _{8}+x _{22}=3a+30d = -21$

If the $n^{th}$ term of an AP be $(2n-1)$, then the sum of its first n terms will be.

1. $n^2-1$

2. $(2n-1)^2$

3. $n^2$

4. $n^2+1$

Correct Option: C
Explanation:
$a _n=(2n-1)$
$\Rightarrow$  $a _1=2\times 1-1=1$
$\Rightarrow$  $a _2=2\times 2-1$
$=4-1$
$=3$
$\Rightarrow$  $d=a _1-a _1=3-1$
$\therefore$  $d=2$
$\Rightarrow$  $S _1=\dfrac{n}{2}[2a _1+(n-1)d]$

$=\dfrac{n}{2}[2(1)+(n-1)2]$

$=\dfrac{n}{2}[2+2n-2]$

$=\dfrac{n}{2}\times 2n$

$=n^2$

If $a,b,c$ are distnct and the roots of $(b-c)x^{2}+(c-a)x+(a-b)=0$are equal, then $a,b,c$ are in

1. Arithmetic progression

2. Geometric prograsson

3. Harmonic prograssiion

4. Arithmetco- Geometric prograssion

Correct Option: A

If the $p^{th}$, $q^{th}$ and $r^{th}$ terms of an A.P. are P, Q, R respectively, then $P(q-r)+Q(r-p)+R(p-q)$ is equal to _________.

1. $0$

2. $1$

3. pqr

4. p$+$qr

Correct Option: A
Explanation:

let $a$  be first term of A.P. and $d$ is it's common difference

$p$ th term is $a+(p-1)d$
$q$ th term is $a+(q-1)d$
$r$ th term is $a+(r-1)d$
therefore $P(q-r) +Q(r-p) +R(p-q)$ on simplifying gives $(q-r +r-p+p-q)+d[(q-r)(p-1)+(q-1)(r-p)+(r-1)(p-q)]$
which on evaluation gives $0$
Hence Option A is correct

If $\sin { \ \alpha },\ \sin ^{ 2 }{ \ \alpha },\ 1,\ \sin ^{ 4 }{ \ \alpha }$ and $\ \sin ^{ 5 }{ \ \alpha }$ are in A.P. where $-\pi <a<\pi$, then $\alpha$ lies in the interval-

1. $\left( \dfrac { -\pi }{ 2 } ,\dfrac { \pi }{ 2 } \right)$

2. $\left( \dfrac { -\pi }{ 3 } ,\dfrac { \pi }{ 3 } \right)$

3. $\left( \dfrac { -\pi }{ 6 } ,\dfrac { \pi }{ 6 } \right)$

4. none of these

Correct Option: A
Explanation:

We have,

$\sin \alpha ,\,{{\sin }^{2}}\alpha ,\,1,\,{{\sin }^{4}}\alpha \,\,and\,\,{{\sin }^{5}}\alpha$ in A.P.

Then,

$\text{First}\,\text{term}\,\,a=\sin \alpha$

$\text{Common}\,\text{difference}\,\,\text{=}\,\text{Second}\,\text{term}\,\text{-}\,\text{first}\,\,\text{term}$

Where

${{T} _{1}}=\,First\,term$

${{T} _{2}}=Second\,term$

${{T} _{3}}=\,Third\,term$

$.......$

Then, we know that,

If the series in an A.P.

${{T} _{2}}-{{T} _{1}}={{T} _{3}}-{{T} _{2}}={{T} _{4}}-{{T} _{3}}={{T} _{5}}-{{T} _{4}}$

${{\sin }^{2}}\alpha -\sin \alpha =1-{{\sin }^{2}}\alpha ={{\sin }^{4}}\alpha -1={{\sin }^{5}}\alpha -{{\sin }^{4}}\alpha$

$\Rightarrow {{\sin }^{2}}\alpha -\sin \alpha =1-{{\sin }^{2}}\alpha$

$\Rightarrow {{\sin }^{2}}\alpha +{{\sin }^{2}}\alpha -\sin \alpha =1$

$\Rightarrow 2{{\sin }^{2}}\alpha -\sin \alpha -1=0$

$\Rightarrow 2{{\sin }^{2}}\alpha -\left( 2-1 \right)\sin \alpha -1=0$

$\Rightarrow 2{{\sin }^{2}}\alpha -2\sin \alpha +\sin \alpha -1=0$

$\Rightarrow 2\sin \alpha \left( \sin \alpha -1 \right)+1\left( \sin \alpha -1 \right)=0$

$\Rightarrow \left( \sin \alpha -1 \right)\left( 2\sin \alpha +1 \right)=0$

$\Rightarrow \sin \alpha -1=0,\,\,\,2\sin \alpha +1=0$

$\Rightarrow \sin \alpha =1,\,\,\,\sin \alpha =\dfrac{-1}{2}$

$\Rightarrow \sin \alpha =\sin \dfrac{\pi }{2},\,\,\,\sin \alpha =-\sin \dfrac{\pi }{6}$

$\Rightarrow \alpha =\dfrac{\pi }{2},\,\,\,\,\sin \alpha =\sin \left( \pi +\dfrac{\pi }{6} \right)\,\,\,\,\,\,\,\,\,\,\because \sin \left( \pi +\theta \right)=-\sin \theta$

$\Rightarrow \alpha =\dfrac{\pi }{2},\,\,\,\,\,\alpha =\dfrac{7\pi }{6}$

Similarly we can show that,

$\alpha =-\dfrac{\pi }{2}$

Hence, $\alpha =\left( -\dfrac{\pi }{2},\,\dfrac{\pi }{2} \right)$

Hence, this is the required answer.

The sum of all the natural numbers from $200$ to $600$(both inclusive) which are neither divisible by $8$ nor by $12$ is?

1. $123968$

2. $133068$

3. $133268$

4. $187332$

Correct Option: C
Explanation:
$D(8)=$ numbers divisible by $8 = 200, 208, 216, 224, 232, 240,.., 592, 600.$
Total $D(8)$ numbers $= 51$
Sum of $D(8)$ numbers $=\left[\dfrac{51}{2}\times (200 + 600)\right] = (51\times 400) = 20400$

$D(12)=$ numbers divisible by $12 = 204, 216, 228, 240, 252, 264,.., 588, 600.$
Total $D(12)$ numbers $=34$
Sum of $D(12)$ numbers $=\left[\dfrac{34}{2}\times (204 + 600)\right] = (17\times 804) = 13668$

Now, $D(8\cap 12) =$ numbers divisible by both $8$ and $12 = 216, 240, 264,..., 576, 600.$

Total $D(8\cap 12)$ numbers $=17$

Sum of $D(8$ intersect $12)$ numbers $= \left[\dfrac{17}{2}\times (216 + 600)\right] = (17\times 408) = 6936$

So, $D(8\cup 12) =$ numbers divisible by either $8$ or $12$

$= D(8) + D(12) - D(8\cap 12).$

So, sum of $D(8\cup 12)$ numbers $= 20400 + 13668 - 6936 = 27132.$

Now, sum of all natural numbers ranging from $200$ to $600 = \left[\dfrac{401}{2}\times (200 + 600)\right] = (401\times 400) = 160400.$

Sum of all natural numbers from $200$ to $600$ which are neither divisible by $8$nor by $12 = (160400 - 27132)$
$= 133268.$

The line joining $A$ $\left( b\cos { \alpha ,\ b\sin { \alpha } } \right)$ and $B$ $\left( a\cos { \beta ,\ a\sin { \beta } } \right)$ is produced to the point $M$ $\left( x,y \right)$, so that $AM$ and $BM$ are in the ration $b:a$. Prove that
$x+y\ \tan { \left( \dfrac { \alpha +\beta }{ 2 } \right) } =0$

1. $-1$

2. $0$

3. $\sin (\alpha + \beta /2)$

4. $\sin (\alpha - \beta /2)$

Correct Option: D
Explanation:
Given $\dfrac{AM}{BM}=\dfrac{b}{a}$
$\Rightarrow M$ divides $AB$ externally in the ratio $b:a$
$\Rightarrow x=\dfrac{ba\cos \beta-ab\cos \alpha}{b-a}$ and $y=\dfrac{ba\sin \beta-ab\sin \alpha}{b-a}$
$\Rightarrow \dfrac{x}{y}=\dfrac{\cos \beta-\cos \alpha}{\sin \beta-\sin \alpha}$
$\cos \beta=\dfrac{1-\tan^2(\beta/2)}{1+\tan^2(\beta/2) }$, $\cos \alpha =\dfrac{1-\tan^2(\alpha /2)}{1+\tan^2(\alpha /2)}$, $\sin \beta=\dfrac{2\tan (\beta /2)}{1+\tan^2(\beta/2)}, \sin \alpha=\dfrac{2\tan \dfrac{\alpha}{2}}{1+\tan^2\dfrac{\alpha}{2}}$
$\Rightarrow \dfrac{x}{y}=\dfrac{\dfrac{1-\tan^2\dfrac{\beta}{2}}{1+\tan^2 \beta/2}-\dfrac{1-\tan^2\dfrac{\alpha}{2}}{1+\tan^2 \alpha/2}}{\dfrac{2\tan \beta/2}{1+\tan^2 \beta/2}-\dfrac{2\tan \alpha/2}{1+\tan^2 \alpha/2}}=\displaystyle \dfrac { 1+\tan ^{ 2 }{ \dfrac { \alpha }{ 2 } } -\tan ^{ 2 }{ \dfrac { \beta }{ 2 } } -\tan ^{ 2 }{ \dfrac { \alpha }{ 2 } } \tan ^{ 2 }{ \dfrac { \beta }{ 2 } } -1-\tan ^{ 2 }{ \dfrac { \beta }{ 2 } } +\tan ^{ 2 }{ \dfrac { \alpha }{ 2 } } +\tan ^{ 2 }{ \dfrac { \alpha }{ 2 } } \tan ^{ 2 }{ \dfrac { \beta }{ 2 } } }{ 2\tan { \dfrac { \beta }{ 2 } } +2\tan { \dfrac { \beta }{ 2 } } \tan ^{ 2 }{ \dfrac { \alpha }{ 2 } } -2\tan { \dfrac { \alpha }{ 2 } } -2\tan { \dfrac { \alpha }{ 2 } \tan ^{ 2 }{ \dfrac { \beta }{ 2 } } } }$
$\Rightarrow \dfrac { x }{ y } =\dfrac { 2\left( \tan ^{ 2 }{ \dfrac { \alpha }{ 2 } } -\tan ^{ 2 } \beta /2 \right) }{ 2\left( \tan \beta /2-\tan \dfrac { \alpha }{ 2 } \right) \left( 1-\tan \dfrac { \alpha }{ 2 } \tan { \beta /2 } \right) } =\dfrac { -\left( \tan \dfrac { \alpha }{ 2 } -\tan \dfrac { \beta }{ 2 } \right) \left( \tan \dfrac { \alpha }{ 2 } +\tan \dfrac { \beta }{ 2 } \right) }{ \left( \tan \dfrac { \alpha }{ 2 } -\tan \dfrac { \beta }{ 2 } \right) \left( 1-\tan \dfrac { \alpha }{ 2 } \tan \dfrac { \beta }{ 2 } \right) }$
$\Rightarrow x+y\dfrac{\tan \dfrac{\alpha}{2}+\tan \beta/2}{1-\tan \dfrac{\alpha}{2}\tan \dfrac{\beta}{2}}=0\Rightarrow x+y\tan \left(\dfrac{\alpha+\beta}{2}\right)=0$ Hence proved

Find the sum of the first $15$ terms of the following sequences having $n$th term as
${a} _{n}=3+4n$

1. 525

2. 563

3. 184

4. 189

Correct Option: A
Explanation:

$a _{n}=3+4n$ (Given)

Now,
$a _{1}=3+4\times 1=7$
$a _{2}=3+4\times 2=11$
$a _{3}=3+4\times 3=15$
So the series is
The sum of first is turns is
$S _{n}=\dfrac{n}{2}[2a+(n-1)d]$
$a=7, n=15, d=4$
$S _{n}=\dfrac{15}{2}[2\times 7+(15-1).4]$
$S _{n}=\dfrac{15}{2}[14+56]$
$S _{n}=\dfrac{15\times 70}{2}$
$S _{n}=15\times 35$
$S _{n}=525$
The sum of first $15$ terms of given series is
$S _{n}=525$

Let ${V} _{r}$ denote the sum of the first $r$ terms of an A.P whose first term is $r$ and common difference is $(2r-1)$.Let

${T} _{r}={V} _{r+1}-{V} _{r}-2$ and

${Q} _{r}={T} _{r+1}-{T} _{r}$ $T$ is always

1. an odd number

2. an even number

3. a prime number

4. a composite number

Correct Option: D
Explanation:

We have sum of $n$ terms $=\dfrac{n}{2}\left(2a+(n-1)d\right)$ where $a$ is the first term, $n$ is the number of terms and $d$ is the common difference in an A.P.
From the passage $n=r,$ $a=2r$ and $d=(2r-1)$
$\therefore {V} _{r}=\dfrac{r}{2}\left[2r+(r-1)(2r-1)\right]$
$\Rightarrow \dfrac{r}{2}\left[2r+2{r}^{2}-3r+1\right]=\frac{r}{2}\left[2{r}^{2}-r+1\right]$
Thus ${V} _{r}=\frac{1}{2}\left[2{r}^{3}-{r}^{2}+r\right]={r}^{3}-\frac{{r}^{2}}{2}+\frac{r}{2}$
Now ${T} _{r}={V} _{r+1}-{V} _{r}-2$
From above ${V} _{r}={r}^{3}-\frac{{r}^{2}}{2}+\frac{r}{2}$
${V} _{r+1}={\left(r+1\right)}^{3}-\frac{{\left(r+1\right)}^{2}}{2}+\frac{\left(r+1\right)}{2}$
We have ${T} _{r}={V} _{r+1}-{V} _{r}-2$
${T} _{r}={r}^{3}-\frac{{r}^{2}}{2}+\frac{r}{2}-\left({\left(r+1\right)}^{3}-\frac{{\left(r+1\right)}^{2}}{2}+\frac{\left(r+1\right)}{2}\right)-2$
On simplifying, we get

${T} _{r}={\left(r+1\right)}^{3}-{r}^{3}-\frac{1}{2}\left({\left(r+1\right)}^{2}-{r}^{2}\right)+\frac{1}{2}\left(r+1-r\right)-2$
$\Rightarrow{T} _{r}=\left(r+1-r\right)\left({\left(r+1\right)}^{2}+r\left(r+1\right)+{r}^{2}\right)+\frac{1}{2}-2$
On simplifying, we get
${T} _{r}={r}^{2}+1+2r+{r}^{2}+r+{r}^{2}+\frac{1}{2}\left(-2r-1+1\right)-2$
${T} _{r}=3{r}^{2}+2r-1=\left(3r-1\right)\left(r+1\right)$
We have ${T} _{1}=\left(3-1\right)\left(1+1\right)=2.2$
${T} _{2}=\left(3\times2-1\right)\left(2+1\right)=5.3$
${T} _{3}=\left(3\times3-1\right)\left(3+1\right)=8.4$ which are in A.P
Thus,${T} _{n}=\left(3n-1\right)\left(n+1\right)$
From the above sequence, we note that
Product of even number and an odd number is Even
Product of odd number and an odd number is odd
Product of even number and an even number is Even
and we see that every term is a composite number.
Hence, their sum is a composite number.
$\therefore T$ is a composite number.

Let $f(x)=3ax^{2}-4bx+c(a,b,c \in R, a \neq 0)$ where $a,b,c$ are in $A.P$. Then the equation $f(x)=0$ has

1. No real solution.

2. Two unequal real roots.

3. Sum of roots always negative.

4. Product of roots always positive.

Correct Option: B
Explanation:

Since $a,b,c$ are in A.P., so,

$2b = a + c$

$4{b^2} = {\left( {a + c} \right)^2}$

The discriminant of the given function$f\left( x \right) = 3a{x^2} - 4bx + c$ is,

$D = 16{b^2} - 12ac$

$= 4{\left( {a + c} \right)^2} - 12ac$

$= 4\left[ {\left( {{a^2} + {c^2} + 2ac} \right) - 3ac} \right]$

$= 4\left( {{a^2} + {c^2} - ac} \right)$

$= 4\left( {{a^2} + {c^2} - 2ac + ac} \right)$

$= 4\left( {{{\left( {a - c} \right)}^2} + ac} \right)$

Case 1: If $a$ and$c$ are of opposite signs, then, $D = \left( + \right){\rm{ve}}$.

Case 2: If $a$ and$c$ are of same signs, then, $D = \left( + \right){\rm{ve}}$.

This shows that $f\left( x \right) = 0$ has two unequal real roots.

If $ab + bc + ca =0$  , then the value of  $\frac{1}{{{a^2} - bc}} + \frac{1}{{{b^2} - ca}} + \frac{1}{{{c^2} - ab}}$ will be

1. -1

2. a+b+c

3. 0

4. ab

Correct Option: B
Explanation:
$\dfrac{1}{a^2-bc}+\dfrac{1}{b^2-ca}+\dfrac{1}{c^2-ab}$
Given $ab+bc+ca=0$
now $-bc=ab+ca$
$-ca=ab+bc$
$-ab=bc+ca$
$\dfrac{1}{a^2+(ab+ca)}+\dfrac{1}{b^2+(ab+bc)}+\dfrac{1}{c^2+(bc+ca)}$
$=\dfrac{1}{a(a+b+c)}+\dfrac{1}{b(a+b+c)}+\dfrac{1}{c(a+b+c)}$
$=\dfrac{bc+ca+ab}{abc(a+b+c)}=0$.

If $x,y,z$ are in A.P. then the value of the det A where $A=\begin{bmatrix} 4 & 5 & 6 & x \ 5 & 6 & 7 & y \ 6 & 7 & 8 & z \ x & y & z & 0 \end{bmatrix},$ is

1. $0$

2. $1$

3. $2$

4. $-1$

Correct Option: A
Explanation:
$|A|=\begin{vmatrix} 4 & 5 & 6 & x \\ 5 & 6 & 7 & y \\ 6 & 7 & 8 & z \\ x & y & z & 0 \end{vmatrix}$
$=-x\begin{vmatrix} 5 & 6 & x \\ 6 & 7 & y \\ 7 & 8 & z \end{vmatrix}+y\begin{vmatrix} 4 & 6 & x \\ 5 & 7 & y \\ 6 & 8 & z \end{vmatrix}-z\begin{vmatrix} 4 & 5 & x \\ 5 & 6 & y \\ 6 & 7 & z \end{vmatrix}+0\begin{vmatrix} 4 & 5 & 6 \\ 5 & 6 & 7 \\ 6 & 7 & 8 \end{vmatrix}$
$=-x(0)+y(0)-z(0)+0$
Determinate value of a matrix is
zero if all of its rows or
column are in A.P.
In all above $3\times 3$ determinate,
each column is A.P.
$\Rightarrow |A|=0$
Choose the correct choice in the given and justify, $11th$ term of the A.P. : $-3,-\dfrac{1}{2},2,...,$ is,
1. $28$

2. $22$

3. $-38$

4. $-48\dfrac{1}{2}$

Correct Option: B
Explanation:

$first\, \, term\, \, a=-3$

$\ common\, \, difference\, \, d=\dfrac { { -1 } }{ 2 } -\left( { -3 } \right)$

$\ =\dfrac { { -1 } }{ 2 } +3=\dfrac { { -1+6 } }{ 2 } =\dfrac { 5 }{ 2 }$

Now,

$\ { a _{ n } }=a+\left( { n-1 } \right) d$

$\ { a _{ n } }=-3+\left( { 11-1 } \right) \times \dfrac { 5 }{ 2 }$

$\ =-3+25$

$\ { a _{ 11 } }=22$

If $\displaystyle \frac{b+c-a}{a},\frac{c+a-b}{b},\frac{a+b-c}{c}$ are in A.P.,then $\displaystyle\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in

1. A.G.P

2. G.P

3. H.P

4. A.P

Correct Option: D
Explanation:

$\displaystyle \frac{b+c-a}{a},\frac{c+a-b}{b},\frac{a+b-c}{c}$ are in $AP$

If each term of a given arithmetic progression be increased, decreased,multiplied or divided by the same non-zero quantity,then the resultant series thus obtained will also be in $AP$.

adding $2$ to each term
$\Rightarrow \displaystyle \frac{b+c-a}{a}+2,\frac{c+a-b}{b}+2,\frac{a+b-c}{c}+2$ are also in $AP$

$\Rightarrow \displaystyle \frac{b+c+a}{a},\frac{c+a+b}{b},\frac{a+b+c}{c}$ are also in $AP$

dividing each term by $a+b+c$

$\therefore\displaystyle \frac{1}{a},\frac{1}{b},\frac{1}{c}$ are also in $AP$
Hence, option D.

The sum of all terms of the arithmetic progression having ten terms except for the first tens, is 99, and except for the sixth term, is 89. Find the third term of the progression if the sum of the first and the fifth term is equal to 10.

1. 15

2. 5

3. 8

4. 10

Correct Option: B
Explanation:

Given:

${ S } _{ 10 }=99+{ T } _{ 1 }..........(i)\ { S } _{ 10 }=89+{ T } _{ 6 }..........(ii)$
where ${ S } _{ 10 }$ is the sum of $10$ terms of the A.P. and ${ T } _{ 1 }, { T } _{ 6 }$ are the first and sixth term respectively.
Say $a$ and $d$ are the first term and common difference of the A.P. respectively.
$\ \therefore { S } _{ 10 }=5\left{ 2a+9d \right} ;\quad { T } _{ 1 }=a;\quad { T } _{ 6 }=a+5d........(iii)\ \therefore 5\left{ 2a+9d \right} =a+99........(iv)\ 5\left{ 2a+9d \right} =a+89+5d........(v)\$
Subtracting (iv) and (v), we get,
$10-5d=0\ =>d=2........(vi)$
Also given that
${ T } _{ 1 }+{ T } _{ 5 }=10\ =>a+a+4d=10\ =>2a+4\times 2=10\ =>2a=2\ =>a=1$
$\therefore { T } _{ 3 }=a+2d=1+2\times 2=5$

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