Tag: maths

Questions Related to maths

The $n^{th}$ term of the series $3+7+14+24+.....$ is

  1. $\dfrac{n-1}{2}(3n+2)$

  2. $3+\dfrac{n-1}{2}(3n+2)$

  3. $1+\dfrac{n-1}{2}(3n+2)$

  4. $3+\dfrac{(n+1)(3n+2)}{2}$


Correct Option: B

A line passes through the variable point $A(\lambda +1,2\lambda)$ meets the lines $7x+y-16=0,\ 5x-y-8=0,\ x-5y+8=0$ at $b,c,d$ respectively. Then $AC, AB, AD$ are in

  1. A.P

  2. G.P

  3. H.P

  4. None of these


Correct Option: A

If $T _n$ denotes the nth terms of the series.$ 2+3+6+11+18+..........$ , then $T _50$ is :

  1. $49^2-1$

  2. $49^2$

  3. $50^2+1$

  4. $49^2+2$


Correct Option: D
Explanation:

$2+3+6+11+18+.................$

$\begin{matrix} { T _{ n } }=a{ x^{ 2 } }+bx+c \ { T _{ 1 } }=a+b+c=2 \ { T _{ 2 } }=a+b+c=3 \ { T _{ 3 } }=a+b+c=6 \ 3a+b=1 \ 5a+b=3 \ 2a=2 \ a=1 \ b=-2 \ c=3 \  \end{matrix}$
${T _n} = {n^2} - 2n + 3$
${T _{50}} = {\left( {50} \right)^2} - 2\left( {50} \right) + 3$
${T _{50}} = {\left( {49} \right)^2} + 2$
Hence,
Option $D$ is correct answer.

If the m+n, n+p, p+n terms of an AP are a, b, c respectively, then m(b-c)+n(c-a)+p(a-b) is

  1. 1

  2. a+b+c

  3. m+n+p

  4. 0


Correct Option: D
Explanation:
$T _{m}=A+(m-1)d=a$
$T _{n}=A+(n-1)d=b$
$T _{p}=A+(p-1)d=c$
From these equations, we get,
$a-b=(m-n)d$
$b-c=(n-p)d$
and $c-a=(p-m)d$
Now,
$m(b-c)+n(c-a)+p(a-b)=m(n-p)d+n(p-m)d+p(m-n)d$
$=d(mn-mp+np-nm+pm-pn)$
$=d \times 0=0$

Let $a,b,c$ be in AP and $k\neq 0$ be a real number. WHich of trhe following are correct ?
1.$ka,kb,kc$ are in Ap
2. $k-a,k-b,k-c$ are in AP
3. $\dfrac{a}{k},\dfrac{b}{k},\dfrac{c}{k}$ are in AP
Select the correct answer using the code given below :

  1. 1 and 2 only

  2. 2 and 3 only

  3. 1 and 3 only

  4. 1, 2 and 3 only


Correct Option: D
Explanation:

$a, b, c \in A.P. \quad \left( \text{Given} \right)$

Therefore,
$(1) \quad\text{when a constant term is multiplied to each term of an A.P. then the resultant is also an A.P.}$
$ka, kb, kc \in A.P., \quad k \ne 0$

$(2)\quad\text{when a constant is subtracted from each term of an A.P. then the resultant is also an A.P.}$
$\Rightarrow k - a, k-b, k - c \in A.P.$

$(3)\quad\text{when a constant is divided from each term of an A.P. then the resultant is also an A.P.}$
$\Rightarrow \cfrac{a}{k}, \cfrac{b}{k}, \cfrac{c}{k} \in A.P.$
Hence all statements are correct.

Find the next term of the series:
$22, 26, 29, 31,$ ..........

  1. $32$

  2. $33$

  3. $34$

  4. $35$


Correct Option: A
Explanation:

Lets find the difference between the numbers of the series:

$22$       $26$       $29$       $31$       $x$.....
      $4$          $3$         $2$        $1$
Since, the difference is in A.P.
$\therefore x=31+1=32$
Hence, the answer is $32$.

$\displaystyle a^{2}\left ( b+c \right ),b^{2}\left ( c+a \right ),c^{2}\left ( a+b \right )$ provided $\displaystyle \sum ab\neq 0.$ if $a=b=c$ then above series is in:

  1. AP

  2. GP

  3. HP

  4. AGP


Correct Option: A,B,C,D

If the $m^{th}$ term and the $n^th$ term of an AP are respectively $\displaystyle \frac { 1 }{ n } $ and $\displaystyle \frac { 1 }{ m } $, then the $mn^{th}$ term of the AP is

  1. $\displaystyle \frac { 1 }{ mn } $

  2. $\displaystyle \frac { m }{ n } $

  3. $\displaystyle 1$

  4. $\displaystyle \frac { n }{ m } $


Correct Option: C
Explanation:

Let a and d be the first term and common difference of an AP.
Since, $\displaystyle { T } _{ m }=\frac { 1 }{ n } $
$\displaystyle \therefore a+\left( m-1 \right) d=\frac { 1 }{ n } $....(i)
and $\displaystyle { T } _{ n }=\frac { 1 }{ m } $
$\displaystyle \Rightarrow a+\left( n-1 \right) d=\frac { 1 }{ m } $.....(ii)
On solving Eqs. (i) and (ii), we get
$\displaystyle a=\frac { 1 }{ mn } and\quad d=\frac { 1 }{ mn } $
$\displaystyle \therefore \quad { T } _{ mn }=a+\left( mn-1 \right) d$


$\displaystyle =\frac { 1 }{ mn } +\frac { \left( mn-1 \right)  }{ mn } $

$\displaystyle =\frac { mn }{ mn } =1$

Which one is an example of A.P. property?

  1. Constant $a$ is added to each term of an A.P. will form a new A.P with different common difference

  2. Constant $a$ is subtracted to each term of an A.P. will form a new A.P with different common difference

  3. Constant $a$ is divided to each term of an A.P. will not form a new A.P with same common difference

  4. Constant $a$ is added to each term of an A.P. will form a new A.P with same common difference


Correct Option: D
Explanation:

If you will subtract or add a constant number to each terms of an AP, then later sequence will also be in AP with same common difference

Hence option 'D' is correct choice 

Identify the property of A.P. used in the sequence: 

$(3 - x), (5 - x), (7 - x), (9 - x)$

  1. $3$ is a constant subtracted from the sequence

  2. $-x$ is a constant subtracted from the sequence

  3. $x$ is a constant subtracted from the sequence

  4. Number is a constant subtracted from the sequence


Correct Option: C
Explanation:
Given sequence is $(3-x),(5-x),(7-x),(9-x)$
Here $3,5,7,9$ are in AP with common difference $2$
Using the property of AP, if we subtract the same no. from each term, then it will remain in AP.
i.e. $3-x , 5-x, 7-x, 9-x$ is all equal to $2$ which will be in AP.