Tag: complete the a.p series with given information

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

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 

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

  1. -1

  2. a+b+c

  3. 0

  4. ab

Show answer
Correct Option: B
Explanation:
$\dfrac{1}{a^2-bc}+\dfrac{1}{b^2-ca}+\dfrac{1}{c^2-ab}$
Given $ab+bc+ca=0$
now $-bc=ab+ca$
$-ca=ab+bc$
$-ab=bc+ca$
$\dfrac{1}{a^2+(ab+ca)}+\dfrac{1}{b^2+(ab+bc)}+\dfrac{1}{c^2+(bc+ca)}$
$=\dfrac{1}{a(a+b+c)}+\dfrac{1}{b(a+b+c)}+\dfrac{1}{c(a+b+c)}$
$=\dfrac{bc+ca+ab}{abc(a+b+c)}=0$.

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 

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. $-1$

Show answer
Correct Option: A
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.
$\Rightarrow |A|=0$
Choose the correct choice in the given and justify, $11th$ term of the A.P. : $-3,-\dfrac{1}{2},2,..., $ is,
maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

  1. $28$

  2. $22$

  3. $-38$

  4. $-48\dfrac{1}{2}$

Show answer
Correct Option: B
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 $

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 

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

  1. A.G.P

  2. G.P

  3. H.P

  4. A.P

Show answer
Correct Option: D
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.

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.

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

  1. 15

  2. 5

  3. 8

  4. 10

Show answer
Correct Option: B
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
${ T } _{ 1 }+{ T } _{ 5 }=10\ =>a+a+4d=10\ =>2a+4\times 2=10\ =>2a=2\ =>a=1$
$\therefore { T } _{ 3 }=a+2d=1+2\times 2=5$