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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 $ab + bc + ca =0$  , then the value of  $\frac{1}{{{a^2} - bc}} + \frac{1}{{{b^2} - ca}} + \frac{1}{{{c^2} - ab}}$ will be 

  1. -1

  2. a+b+c

  3. 0

  4. ab
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\dfrac{1}{a^2-bc}+\dfrac{1}{b^2-ca}+\dfrac{1}{c^2-ab}$
Given $ab+bc+ca=0$
now $-bc=ab+ca$
$-ca=ab+bc$
$-ab=bc+ca$
$\dfrac{1}{a^2+(ab+ca)}+\dfrac{1}{b^2+(ab+bc)}+\dfrac{1}{c^2+(bc+ca)}$
$=\dfrac{1}{a(a+b+c)}+\dfrac{1}{b(a+b+c)}+\dfrac{1}{c(a+b+c)}$
$=\dfrac{bc+ca+ab}{abc(a+b+c)}=0$.
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 in A.P. then the value of the det A where $A=\begin{bmatrix} 4 & 5 & 6 & x \ 5 & 6 & 7 & y \ 6 & 7 & 8 & z \ x & y & z & 0 \end{bmatrix},$ is 

  1. $0$
  2. $1$
  3. $2$
  4. $-1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$|A|=\begin{vmatrix} 4 & 5 & 6 & x \\ 5 & 6 & 7 & y \\ 6 & 7 & 8 & z \\ x & y & z & 0 \end{vmatrix}$
$=-x\begin{vmatrix} 5 & 6 & x \\ 6 & 7 & y \\ 7 & 8 & z \end{vmatrix}+y\begin{vmatrix} 4 & 6 & x \\ 5 & 7 & y \\ 6 & 8 & z \end{vmatrix}-z\begin{vmatrix} 4 & 5 & x \\ 5 & 6 & y \\ 6 & 7 & z \end{vmatrix}+0\begin{vmatrix} 4 & 5 & 6 \\ 5 & 6 & 7 \\ 6 & 7 & 8 \end{vmatrix}$
$=-x(0)+y(0)-z(0)+0$
Determinate value of a matrix is 
zero if all of its rows or
column are in A.P. 
In all above $3\times 3$ determinate,
each column is A.P.
$\Rightarrow |A|=0$
Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap
Choose the correct choice in the given and justify, $11th$ term of the A.P. : $-3,-\dfrac{1}{2},2,..., $ is,
  1. $28$
  2. $22$
  3. $-38$
  4. $-48\dfrac{1}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$first\, \, term\, \, a=-3 $

$\ common\, \, difference\, \, d=\dfrac { { -1 } }{ 2 } -\left( { -3 } \right)$

$  \ =\dfrac { { -1 } }{ 2 } +3=\dfrac { { -1+6 } }{ 2 }  =\dfrac { 5 }{ 2 }$ 

Now,

 $ \ { a _{ n } }=a+\left( { n-1 } \right) d $

$\ { a _{ n } }=-3+\left( { 11-1 } \right) \times \dfrac { 5 }{ 2 }$

$  \ =-3+25$

$ \ { a _{ 11 } }=22 $

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

If $\displaystyle \frac{b+c-a}{a},\frac{c+a-b}{b},\frac{a+b-c}{c}$ are in A.P.,then $\displaystyle\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in 

  1. A.G.P

  2. G.P

  3. H.P

  4. A.P
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

 $\displaystyle \frac{b+c-a}{a},\frac{c+a-b}{b},\frac{a+b-c}{c}$ are in $AP$


If each term of a given arithmetic progression be increased, decreased,multiplied or divided by the same non-zero quantity,then the resultant series thus obtained will also be in $AP$.

adding $2$ to each term
$\Rightarrow \displaystyle \frac{b+c-a}{a}+2,\frac{c+a-b}{b}+2,\frac{a+b-c}{c}+2$ are also in $AP$

$\Rightarrow \displaystyle \frac{b+c+a}{a},\frac{c+a+b}{b},\frac{a+b+c}{c}$ are also in $AP$

dividing each term by $a+b+c$

$\therefore\displaystyle \frac{1}{a},\frac{1}{b},\frac{1}{c}$ are also in $AP$
Hence, option D.

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

The sum of all terms of the arithmetic progression having ten terms except for the first tens, is 99, and except for the sixth term, is 89. Find the third term of the progression if the sum of the first and the fifth term is equal to 10.

  1. 15

  2. 5

  3. 8

  4. 10
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given:

${ S } _{ 10 }=99+{ T } _{ 1 }..........(i)\ { S } _{ 10 }=89+{ T } _{ 6 }..........(ii)$
where ${ S } _{ 10 }$ is the sum of $10$ terms of the A.P. and ${ T } _{ 1 }, { T } _{ 6 }$ are the first and sixth term respectively.
Say $a$ and $d$ are the first term and common difference of the A.P. respectively.
$\ \therefore { S } _{ 10 }=5\left{ 2a+9d \right} ;\quad { T } _{ 1 }=a;\quad { T } _{ 6 }=a+5d........(iii)\ \therefore 5\left{ 2a+9d \right} =a+99........(iv)\ 5\left{ 2a+9d \right} =a+89+5d........(v)\ $
Subtracting (iv) and (v), we get,
$10-5d=0\ =>d=2........(vi)$
Also given that
${ T } _{ 1 }+{ T } _{ 5 }=10\ =>a+a+4d=10\ =>2a+4\times 2=10\ =>2a=2\ =>a=1$
$\therefore { T } _{ 3 }=a+2d=1+2\times 2=5$