Tag: arithmetic progressions

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Questions Related to arithmetic progressions

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 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

  1. Arithmetic progression

  2. Geometric progression

  3. Harmonic progression

  4. Arithmetico-Geometric progression
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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.

  1. True

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

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 the straight lines $ax+by+c=0$ wil always pass through the point ..........

  1. $(1,2)$
  2. $(1,5)$
  3. $(3,2)$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

An AP consists of $15$ terms. The three middle most terms is $69$ and the last three terms is $123$. Find the A.P.

  1. 2,5,8,11.....44

  2. 2,4,8,....62

  3. 2,3,4....16

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The $three$ middle most terms of $15$ terms is $69.$

${a} _{7}+{a} _{8}+{a} _{9}=69$
${a} _{13}+{a} _{14}+{a} _{15}=123$
$a+6d+a+7d+a+8d=69\Longrightarrow 3a+21d=69$
$a+12d+a+13d+a+14d=123\Longrightarrow 3a+39d=123$

$3a+21d=69$
$3a+39d=123$

          $18d=54,\d=3,\a=2.$
$2,5,8,11,\dots 44.$  

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

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 

  1. $1$
  2. $0$
  3. $xyz$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
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}$
$=1$

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

The largest term to common to the sequences $1,11,21,31,... to 100$ terms and $31,36,41,46, ...to 100$ terms is 

  1. $511$
  2. $471$
  3. $281$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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 =$20\times 25\times 11=511$
Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

If $a^{2},b^{2},c^{2}$ are in $AP$, then which of the following are in $AP$?

  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. $\dfrac{1}{a^{2}},\dfrac{1}{a^{2}},\dfrac{1}{c^{2}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.
Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

If $a,b$ and $c$ are in $A.P.,$ then $\dfrac{(a-c)^{2}}{b^{2}-ac}=$ ?

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 } $
$\cfrac { { a }^{ 2 }+{ c }^{ 2 }-2ac }{ { \left( \cfrac { a+c }{ 2 }  \right)  }^{ 2 }-ac } =\cfrac { { a }^{ 2 }+{ c }^{ 2 }-2ac }{ \left( \cfrac { { a }^{ 2 }+{ c }^{ 2 }-2ac }{ 2 }  \right)  } =2$

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 $\left|\begin{matrix} x+1 & x+2 & x+a \ x+2 & x+3 & x+b \ x+3 & x+4 & x+c \end{matrix}\right|$

  1. $1$
  2. $0$
  3. $-1$
  4. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
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