Tag: complete the a.p series with given information

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

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. $\left[ {6,12} \right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\infty$

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

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

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

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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. $8:2013$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If A, B, C are in AP, then 2B = A+C. Since A+B+C = 180, B=60. Using the cosine rule b^2 = a^2 + c^2 - 2ac cos(B), and cos(60)=1/2, we get b^2 = a^2 + c^2 - ac.

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$a,b,c$ are in A.P
$\Rightarrow\,b-a=c-b$
$\Rightarrow\,2b=a+c$
${a}^{3}+{c}^{3}-8{b}^{3}$
$={a}^{3}+{c}^{3}-{\left(2b\right)}^{3}$
$={a}^{3}+{c}^{3}-{\left(a+c\right)}^{3}$
$={a}^{3}+{c}^{3}-{a}^{3}-{c}^{3}-3ac\left(a+c\right)$
$=-3ac\left(2b\right)=-6abc$

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 }\  $ 
$\therefore { \left( b-c \right)  }^{ 2 },{ \left( c-a \right)  }^{ 2 },{ \left( a-b \right)  }^{ 2 }\quad are\quad in\quad AP$

Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on 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. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $2$

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

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. $42.5$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 
$Mean _{correct}=\dfrac{800+20}{20}=41$

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

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. ${ log } _{ 5 }2$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 } $

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given 1/a, 1/b, 1/c are in AP, 2/b = 1/a + 1/c. Substituting this into the expression and simplifying leads to the result 4/ac - 3/b^2.