Tag: maths

Questions Related to maths

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

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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

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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

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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
${ T } _{ 1 }+{ T } _{ 5 }=10\ =>a+a+4d=10\ =>2a+4\times 2=10\ =>2a=2\ =>a=1$
$\therefore { T } _{ 3 }=a+2d=1+2\times 2=5$

Find the side of the square whose diagonal is $16 \sqrt 2$ cm.

maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

  1. $4$ cm

  2. $16$ cm

  3. $8$ cm

  4. $16\sqrt 2$ cm

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Correct Option: B
Explanation:

We know that,

1) All angles of a square are congruent. i.e $90^o$
2) Diagonal of a square bisects each of its angles.
Therefore, the square gets divided into $2$ triangles of degrees $45^o-45^o-90^o$
$\therefore \sin 45^o = \cfrac {\text {side}}{\text {hyp}}$ 
$\therefore \cfrac {1}{\sqrt 2} = \cfrac {\text {side}}{16 \sqrt 2}$
$\therefore$ side of the square $= 16$ cm.

The sides of triangle are in A.P. and the greatest angle exceeds the least by 90. The sides are in the ratio _____________.

maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

  1. $1 : 2 : \sqrt { 2 }$

  2. $1 : \sqrt { 3 } : 2$

  3. $\sqrt { 7 } + 1 : \sqrt { 7 } : \sqrt { 7 } - 1$

  4. $\sqrt { 3 } + 1 : 1 : \sqrt { 3 } - 1$

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Correct Option: C

If  H is orthocenter of triangle PQR then PH + QH + RH is 

maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

  1. QR cot P + PR cot Q + PQ cot R

  2. (pq + QR + RP) (cot P + cot Q + copt R)

  3. $\dfrac{1}{2r}(cot P + cotQ + cot R)$

  4. none of these

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Correct Option: A

ABC is a triangle right angle at B. D is a point on AC such that $\angle ABD = 45^0$. If AC =$6$ and AD =$2$ , then AB is 

maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

  1. $\dfrac{6}{\sqrt{5}}$

  2. ${3}{\sqrt{2}}$

  3. $\dfrac{12}{\sqrt{5}}$

  4. $2$

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Correct Option: A

consider a triangle PQR in which the relation $ QR^2+PR^2=5*PQ^2$ holds. let G be the point of intersection of the medians PM and QN . then angle QGM is always

maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

  1. less then 45 degree

  2. obtuse

  3. a right angle

  4. acute and larger than 45 degree

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Correct Option: A

If in a $\Delta ABC,\sin A=\sin^{2} B$ and $2\cos^{2}A=3\cos^{2}B$, then the $\Delta ABC$ is 

maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

  1. Right angled

  2. Obtuse angled

  3. Isosceles

  4. Equilateral

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Correct Option: A

Which of the following  can be the sides of a right-angled triangle?

maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

  1. $0.5cm, 1.2 cm, 1.3cm$

  2. $2.4cm, 3.2 cm, 7.9cm$

  3. $5.0cm, 5.25 cm, 7.25cm$

  4. $1.6cm, 3.0 cm, 3.4cm$

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Correct Option: A