Mathematics

Advanced Algebra and Calculus

138 Questions

Advanced algebra and calculus topics cover matrices, complex numbers, infinite geometric series, and differential equations. These mathematical concepts frequently appear in officer-level aptitude tests. Solving these questions builds a strong foundation for advanced problem solving.

Complex numbersMatrix operationsInfinite geometric seriesDifferential calculusAlgebraic identities

Advanced Algebra and Calculus Questions

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

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

  1. A.P.

  2. H.P

  3. G.P

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

Given,

$\dfrac{b+c-a}{a},\dfrac{c+a-b}{b},\dfrac{a+b-c}{c}$ one in A.P

Now,

$\because \dfrac{b+c-a}{a},\dfrac{c+a-b}{b},\dfrac{a+b-c}{c}$ are in A.P


$\therefore  \dfrac{b+c-a}{a}+2,\dfrac{c+a-b}{b}+2,\dfrac{a+b-c}{c}+2$, must be  in A.P


$\therefore \dfrac{b+c-a+2a}{a},\dfrac{c+a-b+2b}{b},\dfrac{a+b-c+2c}{c}$ are in A.P


$\therefore \dfrac{a+b+c}{a},\dfrac{a+b+c}{b},\dfrac{a+b+c}{c}$ are in A.P


$\because \dfrac{a+b+c}{a},\dfrac{a+b+c}{b},\dfrac{a+b+c}{c}$ are in A.P


$\therefore \dfrac{1}{(a+b+c)}\times \dfrac{(a+b+c)}{a},\dfrac{1}{(a+b+c)}\times \dfrac{(a+b+c)}{b},\dfrac{1}{(a+b+c)}\times \dfrac{(a+b+c)}{c}$ are in A.P


$\therefore \dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ are in A.P
 

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 geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

$\displaystyle \frac{1}{c},(\frac{1}{ca})^{\dfrac{1}{2}},\frac{1}{a}$ is in

  1. AP

  2. GP

  3. HP

  4. NONE
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given series


$\dfrac{1}{c},\left(\dfrac{1}{ca}\right)^{\dfrac{1}{2}},\dfrac{1}{a}$

Lets consider a G.P of elements $A,B,C$

 $\therefore$ Geo.mean $\Rightarrow B^2=AC$

Comparing it with given series.

$A=\dfrac{1}{c}B=\left(\dfrac{1}{ca}\right)^{\dfrac{1}{2}},C=\dfrac{1}{a}$

$\therefore B^2=\left(\dfrac{1}{ca}\right)^{\dfrac{1}{2}\times 2}$

            $=\dfrac{1}{ca}$........(1)

$AC=\dfrac{1}{c}\times \dfrac{1}{a}=\dfrac{1}{ca}$..............(ii)

$\therefore (i)=(ii)$

$\therefore B^2=AC$ So given series is in G.P