Tag: complete the a.p series with given information

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

If $a,b,c$ are distinct and the roots of $\left( b-c \right) { x }^{ 2 }+\left( c-a \right) x+\left( a-b \right) =0$ 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. Arithmetic progression

  2. Geometric progression

  3. Harmonic progression

  4. Arithmetico-Geometric progression

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Correct Option: A
Explanation:

Clearly $x=1$ is a solution
$\therefore$  product of the roots $=\dfrac { a-b }{ b-c }$ 
$\therefore \left( 1 \right) \left( 1 \right) =\dfrac { a-b }{ b-c }$ 
$\Longrightarrow b-c=a-b$
$\Longrightarrow2b=a+c\Longrightarrow a,b,c$ are in Arithmetic progression.

Say true or false.
In an $A.P$., sum of terms equidistant from the beginning and end is constant and is equal to the sum of the first and last term.

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

  1. True

  2. False

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Correct Option: A
Explanation:

Consider mth term of an AP 
$t _m=a+(m-1)d$
Now, consider (n-m)th term:
$t _{n-m}=a+(n-m-1)d$
Sum $= 2a+(n-1)d $

$= a +a+(n-1)d $
$= a+l$
Therefore, it is true that Sum of the terms is equal to the sum of the first and last terms.

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