Tag: maths

Questions Related to maths

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

maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

  1. $(1,2)$

  2. $(1,5)$

  3. $(3,2)$

  4. None of these

Show answer
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.

maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

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

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

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

  4. none

Show answer
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 

maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

  1. $1$

  2. $0$

  3. $xyz$

  4. $None\ of\ these$

Show answer
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 $f _2$

maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

  1. $0$

  2. $1$

  3. $-1$

  4. $3$

Show answer
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 

maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

  1. $511$

  2. $471$

  3. $281$

  4. None of these

Show answer
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 $AP$?

maths arithmetic progressions complete the a.p series with given information properties of an ap problems on 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. $\dfrac{1}{a^{2}},\dfrac{1}{a^{2}},\dfrac{1}{c^{2}}$

Show answer
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 $\dfrac{(a-c)^{2}}{b^{2}-ac}=$ ?

maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
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 $\left|\begin{matrix} x+1 & x+2 & x+a \ x+2 & x+3 & x+b \ x+3 & x+4 & x+c \end{matrix}\right|$

maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

  1. $1$

  2. $0$

  3. $-1$

  4. $2$

Show answer
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

maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

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

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

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

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

Show answer
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 $S(100)<100$. 

maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

  1. True

  2. False

Show answer
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,
$S(100)<100$

Hence, this is the answer.