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

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. $\sum\limits _{r = 1}^{100} {({a _r} + {b _r})} = 10000$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If a_n and b_n are APs, their sum a_n + b_n is also an AP. The common difference of the sum is the sum of the individual common differences.

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

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
Reveal answer Fill a bubble to check yourself
C Correct answer
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
$\quad p\left( 1 \right) ={ 1 }^{ 3 }-3\times { 1 }^{ 2 }+k+4=0\ \Rightarrow k+2=0\ \Rightarrow k=-2$
$\left| k \right| =2$

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

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. $y ^ { 2 } = \frac { 1 } { x 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $(1)$
$\Rightarrow 2y=x+2\ \Rightarrow 2y=2x\ \Rightarrow y=x\ x=y=z$

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

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. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
D Correct answer
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) $

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

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

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. $6$
Reveal answer Fill a bubble to check yourself
D Correct answer
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, $a _4h _7=\cfrac{7}{3}\times \cfrac{18}{7}=6$

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 $1,\,{\log _y}x,\,{\log _z}y,\, - \,15{\log _{x}z}$ 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
Reveal answer Fill a bubble to check yourself
E Correct answer
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

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

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

T_n = a + (n-1)d. T_3 = a + 2d = 8 and T_7 = a + 6d = 24. Subtracting gives 4d = 16, so d = 4. Then a + 8 = 8, so a = 0. T_n = 0 + (n-1)4 = 4n - 4. If the question asks for T_1, it is 0; if it asks for T_0 (not standard), it is -4.