Tag: complete the a.p series with given information

Questions Related to complete the a.p series with given information

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.

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

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

  1. $2013:1$

  2. $1:2013$

  3. $2013:8$

  4. $8:2013$

Show answer
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:-

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

  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

Show answer
Correct Option: B

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

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

  1. $2 a b c$

  2. -$6 a b c$

  3. $4 a b c$

  4. none of these

Show answer
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 ?

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

  1. GP

  2. AP

  3. HP

  4. None of these

Show answer
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 

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:

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:

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

  1. $41$

  2. $49$

  3. $40.5$

  4. $42.5$

Show answer
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 :

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

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

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

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

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

Show answer
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: 

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

  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

Show answer
Correct Option: A