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

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

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

  1. $4$ cm
  2. $16$ cm
  3. $8$ cm
  4. $16\sqrt 2$ cm
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. $1 : 2 : \sqrt { 2 }$
  2. $1 : \sqrt { 3 } : 2$
  3. $\sqrt { 7 } + 1 : \sqrt { 7 } : \sqrt { 7 } - 1$
  4. $\sqrt { 3 } + 1 : 1 : \sqrt { 3 } - 1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the sides be a-d, a, a+d. Using the law of cosines and the condition that the largest angle exceeds the smallest by 90 degrees, one can derive the ratio of the sides as sqrt(7)+1 : sqrt(7) : sqrt(7)-1.

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

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In any triangle, the distance from the orthocenter to the vertices is given by 2R cos A, 2R cos B, and 2R cos C. Summing these and relating them to the side lengths and cotangents leads to the identity PH + QH + RH = QR cot P + PR cot Q + PQ cot R.

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

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 

  1. $\dfrac{6}{\sqrt{5}}$
  2. ${3}{\sqrt{2}}$
  3. $\dfrac{12}{\sqrt{5}}$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the area of triangle ABC as the sum of areas of ABD and BCD, or using trigonometry in right triangles, we find AB = 6/sqrt(5).

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

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

  1. less then 45 degree

  2. obtuse

  3. a right angle

  4. acute and larger than 45 degree
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Apollonius theorem and the properties of medians, the condition QR^2 + PR^2 = 5PQ^2 implies specific geometric constraints on the triangle, leading to the angle QGM being less than 45 degrees.

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

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

  1. Right angled

  2. Obtuse angled

  3. Isosceles

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

Given sin A = sin^2 B and 2 cos^2 A = 3 cos^2 B, substituting cos^2 A = 1 - sin^2 A = 1 - sin^4 B into the second equation allows solving for sin^2 B, which leads to A = 90 degrees.

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

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

  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. $1.6cm, 3.0 cm, 3.4cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A right-angled triangle must satisfy the Pythagorean theorem: a^2 + b^2 = c^2. For 0.5, 1.2, 1.3: 0.25 + 1.44 = 1.69, which is 1.3^2.