Tag: complete the a.p series with given information

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

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

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

  1. will be in A.P.

  2. will be in H.P.

  3. will be in G.P.

  4. $a+2b=c$

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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 ---------.

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

  1. 16 : 7

  2. 7 : 16

  3. 74 : 169

  4. None of the above

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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,

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

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

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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

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

  1. 0

  2. 1

  3. 2

  4. 3

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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

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

  1. $x = y = z$

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

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

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

Show answer
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 

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

  1. $A.P$

  2. $H.P$

  3. $G.P$

  4. $None\ of\ these$

Show answer
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) $

Hence, this is the answer

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

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

  1. AP

  2. GP

  3. HP

  4. none of these

Show answer
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?

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

  1. $2$

  2. $3$

  3. $5$

  4. $6$

Show answer
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, $a _4h _7=\cfrac{7}{3}\times \cfrac{18}{7}=6$

Hence, this is the answer.

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

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

  1. 12

  2. 15

  3. 17

  4. 19

Show answer
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  

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

  1. ${z^3} = x$

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

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

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

  5. All the above

Show answer
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

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