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

$x _{1}, x _{2}, x _{3}, ....$ are in A.P.
If $x _{1} + x _{7} + x _{10} = -6$ and $x _{3} + x _{8} + x _{12} = -11$, then $x _{3} + x _{8} + x _{22} = ?$

  1. $-21$
  2. $-15$
  3. $-18$
  4. $-31$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the first term be $a$ and common difference be $d$

$x _{1}+x _{7}+x _{10}=-6$
$\Rightarrow (a)+(a+6d)+(a+9d)=-6$
$\Rightarrow 3a+15d=-6 \rightarrow \text{eqn}(1) $

$x _{3}+x _{8}+x _{12}=-11$
$\Rightarrow (a+2d)+(a+7d)+(a+11d)=-11$
$\Rightarrow 3a+20d=-11 \rightarrow \text{eqn}(2)$
Solving eqn (1) and (2), we get $a=3$ and $d=-1$

Now, $x _{3}+x _{8}+x _{22}=3a+30d = -21$

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

If the $n^{th}$ term of an AP be $(2n-1)$, then the sum of its first n terms will be.

  1. $n^2-1$
  2. $(2n-1)^2$
  3. $n^2$
  4. $n^2+1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$a _n=(2n-1)$
$\Rightarrow$  $a _1=2\times 1-1=1$
$\Rightarrow$  $a _2=2\times 2-1$
            $=4-1$
            $=3$
$\Rightarrow$  $d=a _1-a _1=3-1$
$\therefore$  $d=2$
$\Rightarrow$  $S _1=\dfrac{n}{2}[2a _1+(n-1)d]$
           
            $=\dfrac{n}{2}[2(1)+(n-1)2]$

            $=\dfrac{n}{2}[2+2n-2]$

            $=\dfrac{n}{2}\times 2n$

            $=n^2$
Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

If $a,b,c$ are distnct and the roots of $(b-c)x^{2}+(c-a)x+(a-b)=0 $are equal, then $a,b,c$ are in

  1. Arithmetic progression

  2. Geometric prograsson

  3. Harmonic prograssiion

  4. Arithmetco- Geometric prograssion
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a quadratic equation to have equal roots, its discriminant must be zero. Setting (c-a)^2 - 4(b-c)(a-b) = 0 leads to (c-a)^2 + 4(b-c)(b-a) = 0, which simplifies to (a+c-2b)^2 = 0, implying a+c = 2b, which is the definition of an arithmetic progression.

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

If the $p^{th}$, $q^{th}$ and $r^{th}$ terms of an A.P. are P, Q, R respectively, then $P(q-r)+Q(r-p)+R(p-q)$ is equal to _________.

  1. $0$
  2. $1$

  3. pqr
  4. p$+$qr
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

let $a$  be first term of A.P. and $ d$ is it's common difference

$ p $ th term is $ a+(p-1)d$
$ q $ th term is $ a+(q-1)d$
$ r $ th term is $ a+(r-1)d$
therefore $ P(q-r) +Q(r-p) +R(p-q)$ on simplifying gives $ (q-r +r-p+p-q)+d[(q-r)(p-1)+(q-1)(r-p)+(r-1)(p-q)]$
which on evaluation gives $0$   
Hence Option A is correct

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

If $\sin { \ \alpha  },\ \sin ^{ 2 }{ \ \alpha  },\ 1,\ \sin ^{ 4 }{ \ \alpha  }$ and $\ \sin ^{ 5 }{ \ \alpha  }$ are in A.P. where $-\pi <a<\pi$, then $\alpha$ lies in the interval-

  1. $\left( \dfrac { -\pi }{ 2 } ,\dfrac { \pi }{ 2 } \right)$
  2. $\left( \dfrac { -\pi }{ 3 } ,\dfrac { \pi }{ 3 } \right)$
  3. $\left( \dfrac { -\pi }{ 6 } ,\dfrac { \pi }{ 6 } \right)$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have,

$\sin \alpha ,\,{{\sin }^{2}}\alpha ,\,1,\,{{\sin }^{4}}\alpha \,\,and\,\,{{\sin }^{5}}\alpha $ in A.P.

Then,

$ \text{First}\,\text{term}\,\,a=\sin \alpha  $

$ \text{Common}\,\text{difference}\,\,\text{=}\,\text{Second}\,\text{term}\,\text{-}\,\text{first}\,\,\text{term} $

Where

$ {{T} _{1}}=\,First\,term $

$ {{T} _{2}}=Second\,term $

$ {{T} _{3}}=\,Third\,term $

$ ....... $

Then, we know that,

If the series in an A.P.

$ {{T} _{2}}-{{T} _{1}}={{T} _{3}}-{{T} _{2}}={{T} _{4}}-{{T} _{3}}={{T} _{5}}-{{T} _{4}} $

$ {{\sin }^{2}}\alpha -\sin \alpha =1-{{\sin }^{2}}\alpha ={{\sin }^{4}}\alpha -1={{\sin }^{5}}\alpha -{{\sin }^{4}}\alpha  $

$ \Rightarrow {{\sin }^{2}}\alpha -\sin \alpha =1-{{\sin }^{2}}\alpha  $

$ \Rightarrow {{\sin }^{2}}\alpha +{{\sin }^{2}}\alpha -\sin \alpha =1 $

$ \Rightarrow 2{{\sin }^{2}}\alpha -\sin \alpha -1=0 $

$ \Rightarrow 2{{\sin }^{2}}\alpha -\left( 2-1 \right)\sin \alpha -1=0 $

$ \Rightarrow 2{{\sin }^{2}}\alpha -2\sin \alpha +\sin \alpha -1=0 $

$ \Rightarrow 2\sin \alpha \left( \sin \alpha -1 \right)+1\left( \sin \alpha -1 \right)=0 $

$ \Rightarrow \left( \sin \alpha -1 \right)\left( 2\sin \alpha +1 \right)=0 $

$ \Rightarrow \sin \alpha -1=0,\,\,\,2\sin \alpha +1=0 $

$ \Rightarrow \sin \alpha =1,\,\,\,\sin \alpha =\dfrac{-1}{2} $

$ \Rightarrow \sin \alpha =\sin \dfrac{\pi }{2},\,\,\,\sin \alpha =-\sin \dfrac{\pi }{6} $

$ \Rightarrow \alpha =\dfrac{\pi }{2},\,\,\,\,\sin \alpha =\sin \left( \pi +\dfrac{\pi }{6} \right)\,\,\,\,\,\,\,\,\,\,\because \sin \left( \pi +\theta  \right)=-\sin \theta  $

$ \Rightarrow \alpha =\dfrac{\pi }{2},\,\,\,\,\,\alpha =\dfrac{7\pi }{6} $

Similarly we can show that,

$\alpha =-\dfrac{\pi }{2}$

Hence, $\alpha =\left( -\dfrac{\pi }{2},\,\dfrac{\pi }{2} \right)$

Hence, this is the required answer.

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

The sum of all the natural numbers from $200$ to $600$(both inclusive) which are neither divisible by $8$ nor by $12$ is?

  1. $123968$
  2. $133068$
  3. $133268$
  4. $187332$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$D(8)=$ numbers divisible by $8 = 200, 208, 216, 224, 232, 240,.., 592, 600.$
Total $D(8)$ numbers $= 51$
Sum of $D(8)$ numbers $=\left[\dfrac{51}{2}\times (200 + 600)\right] = (51\times 400) = 20400$
 
$D(12)=$ numbers divisible by $12 = 204, 216, 228, 240, 252, 264,.., 588, 600.$
Total $D(12)$ numbers $=34$
Sum of $D(12)$ numbers $=\left[\dfrac{34}{2}\times (204 + 600)\right] = (17\times 804) = 13668$

Now, $D(8\cap 12) =$ numbers divisible by both $8$ and $12 = 216, 240, 264,..., 576, 600.$

Total $D(8\cap 12)$ numbers $=17$

Sum of $D(8$ intersect $12)$ numbers $= \left[\dfrac{17}{2}\times (216 + 600)\right] = (17\times 408) = 6936$

So, $D(8\cup 12) =$ numbers divisible by either $8$ or $12$

                         $= D(8) + D(12) - D(8\cap 12).$

So, sum of $D(8\cup 12)$ numbers $= 20400 + 13668 - 6936 = 27132.$

Now, sum of all natural numbers ranging from $200$ to $600 = \left[\dfrac{401}{2}\times (200 + 600)\right] = (401\times 400) = 160400.$

Sum of all natural numbers from $200$ to $600$ which are neither divisible by $8 $nor by $12 = (160400 - 27132)$
                                                                                                                                                $ = 133268.$
Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

The line joining $A$ $\left( b\cos { \alpha ,\ b\sin { \alpha  }  }  \right)$ and $B$ $\left( a\cos { \beta ,\ a\sin { \beta  }  }  \right)$ is produced to the point $M$ $\left( x,y \right)$, so that $AM$ and $BM$ are in the ration $b:a$. Prove that
$x+y\ \tan { \left( \dfrac { \alpha +\beta  }{ 2 }  \right)  } =0$

  1. $-1$
  2. $0$
  3. $\sin (\alpha + \beta /2)$
  4. $\sin (\alpha - \beta /2)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given $\dfrac{AM}{BM}=\dfrac{b}{a}$
$\Rightarrow M$ divides $AB$ externally in the ratio $b:a$
$\Rightarrow x=\dfrac{ba\cos \beta-ab\cos \alpha}{b-a}$ and $y=\dfrac{ba\sin \beta-ab\sin \alpha}{b-a}$
$\Rightarrow \dfrac{x}{y}=\dfrac{\cos \beta-\cos \alpha}{\sin \beta-\sin \alpha}$
$\cos \beta=\dfrac{1-\tan^2(\beta/2)}{1+\tan^2(\beta/2) }$, $\cos \alpha =\dfrac{1-\tan^2(\alpha /2)}{1+\tan^2(\alpha /2)}$, $\sin \beta=\dfrac{2\tan (\beta /2)}{1+\tan^2(\beta/2)}, \sin \alpha=\dfrac{2\tan \dfrac{\alpha}{2}}{1+\tan^2\dfrac{\alpha}{2}}$
$\Rightarrow \dfrac{x}{y}=\dfrac{\dfrac{1-\tan^2\dfrac{\beta}{2}}{1+\tan^2 \beta/2}-\dfrac{1-\tan^2\dfrac{\alpha}{2}}{1+\tan^2 \alpha/2}}{\dfrac{2\tan \beta/2}{1+\tan^2 \beta/2}-\dfrac{2\tan \alpha/2}{1+\tan^2 \alpha/2}}=\displaystyle \dfrac { 1+\tan ^{ 2 }{ \dfrac { \alpha  }{ 2 }  } -\tan ^{ 2 }{ \dfrac { \beta  }{ 2 }  } -\tan ^{ 2 }{ \dfrac { \alpha  }{ 2 }  } \tan ^{ 2 }{ \dfrac { \beta  }{ 2 }  } -1-\tan ^{ 2 }{ \dfrac { \beta  }{ 2 }  } +\tan ^{ 2 }{ \dfrac { \alpha  }{ 2 }  } +\tan ^{ 2 }{ \dfrac { \alpha  }{ 2 }  } \tan ^{ 2 }{ \dfrac { \beta  }{ 2 }  }  }{ 2\tan { \dfrac { \beta  }{ 2 }  } +2\tan { \dfrac { \beta  }{ 2 }  } \tan ^{ 2 }{ \dfrac { \alpha  }{ 2 }  } -2\tan { \dfrac { \alpha  }{ 2 }  } -2\tan { \dfrac { \alpha  }{ 2 } \tan ^{ 2 }{ \dfrac { \beta  }{ 2 }  }  }  } $
$\Rightarrow \dfrac { x }{ y } =\dfrac { 2\left( \tan ^{ 2 }{ \dfrac { \alpha  }{ 2 }  } -\tan ^{ 2 } \beta /2 \right)  }{ 2\left( \tan  \beta /2-\tan  \dfrac { \alpha  }{ 2 }  \right) \left( 1-\tan  \dfrac { \alpha  }{ 2 } \tan { \beta /2 }  \right)  } =\dfrac { -\left( \tan  \dfrac { \alpha  }{ 2 } -\tan  \dfrac { \beta  }{ 2 }  \right) \left( \tan  \dfrac { \alpha  }{ 2 } +\tan  \dfrac { \beta  }{ 2 }  \right)  }{ \left( \tan  \dfrac { \alpha  }{ 2 } -\tan  \dfrac { \beta  }{ 2 }  \right) \left( 1-\tan  \dfrac { \alpha  }{ 2 } \tan  \dfrac { \beta  }{ 2 }  \right)  } $
$\Rightarrow x+y\dfrac{\tan \dfrac{\alpha}{2}+\tan \beta/2}{1-\tan \dfrac{\alpha}{2}\tan \dfrac{\beta}{2}}=0\Rightarrow x+y\tan \left(\dfrac{\alpha+\beta}{2}\right)=0$ Hence proved
Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

Find the sum of the first $15$ terms of the following sequences having $n$th term as
${a} _{n}=3+4n$

  1. 525

  2. 563

  3. 184

  4. 189
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$a _{n}=3+4n$ (Given)

Now, 
$a _{1}=3+4\times 1=7$
$a _{2}=3+4\times 2=11$
$a _{3}=3+4\times 3=15$
So the series is
The sum of first is turns is
$S _{n}=\dfrac{n}{2}[2a+(n-1)d]$
$a=7, n=15, d=4$
$S _{n}=\dfrac{15}{2}[2\times 7+(15-1).4]$
$S _{n}=\dfrac{15}{2}[14+56]$
$S _{n}=\dfrac{15\times 70}{2}$
$S _{n}=15\times 35$
$S _{n}=525$
The sum of first $15$ terms of given series is 
$S _{n}=525$

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

Let ${V} _{r}$ denote the sum of the first $r$ terms of an A.P whose first term is $r$ and common difference is $(2r-1)$.Let

${T} _{r}={V} _{r+1}-{V} _{r}-2$ and 

${Q} _{r}={T} _{r+1}-{T} _{r}$ $T$ is always

  1. an odd number

  2. an even number

  3. a prime number

  4. a composite number
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have sum of $n$ terms $=\dfrac{n}{2}\left(2a+(n-1)d\right)$ where $a$ is the first term, $n$ is the number of terms and $d$ is the common difference in an A.P.
From the passage $n=r,$ $a=2r$ and $d=(2r-1)$
$\therefore {V} _{r}=\dfrac{r}{2}\left[2r+(r-1)(2r-1)\right]$
$\Rightarrow \dfrac{r}{2}\left[2r+2{r}^{2}-3r+1\right]=\frac{r}{2}\left[2{r}^{2}-r+1\right]$
Thus ${V} _{r}=\frac{1}{2}\left[2{r}^{3}-{r}^{2}+r\right]={r}^{3}-\frac{{r}^{2}}{2}+\frac{r}{2}$
Now ${T} _{r}={V} _{r+1}-{V} _{r}-2$
From above ${V} _{r}={r}^{3}-\frac{{r}^{2}}{2}+\frac{r}{2}$ 
${V} _{r+1}={\left(r+1\right)}^{3}-\frac{{\left(r+1\right)}^{2}}{2}+\frac{\left(r+1\right)}{2}$
We have ${T} _{r}={V} _{r+1}-{V} _{r}-2$
${T} _{r}={r}^{3}-\frac{{r}^{2}}{2}+\frac{r}{2}-\left({\left(r+1\right)}^{3}-\frac{{\left(r+1\right)}^{2}}{2}+\frac{\left(r+1\right)}{2}\right)-2$
On simplifying, we get

${T} _{r}={\left(r+1\right)}^{3}-{r}^{3}-\frac{1}{2}\left({\left(r+1\right)}^{2}-{r}^{2}\right)+\frac{1}{2}\left(r+1-r\right)-2$
$\Rightarrow{T} _{r}=\left(r+1-r\right)\left({\left(r+1\right)}^{2}+r\left(r+1\right)+{r}^{2}\right)+\frac{1}{2}-2$
On simplifying, we get
${T} _{r}={r}^{2}+1+2r+{r}^{2}+r+{r}^{2}+\frac{1}{2}\left(-2r-1+1\right)-2$
${T} _{r}=3{r}^{2}+2r-1=\left(3r-1\right)\left(r+1\right)$
We have ${T} _{1}=\left(3-1\right)\left(1+1\right)=2.2$
${T} _{2}=\left(3\times2-1\right)\left(2+1\right)=5.3$
${T} _{3}=\left(3\times3-1\right)\left(3+1\right)=8.4$ which are in A.P
Thus,${T} _{n}=\left(3n-1\right)\left(n+1\right)$
From the above sequence, we note that
Product of even number and an odd number is Even
Product of odd number and an odd number is odd
Product of even number and an even number is Even
and we see that every term is a composite number.
Hence, their sum is a composite number.
$\therefore T$ is a composite number.

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

Let $f(x)=3ax^{2}-4bx+c(a,b,c \in R, a \neq 0)$ where $a,b,c$ are in $A.P$. Then the equation $f(x)=0$ has

  1. No real solution.

  2. Two unequal real roots.

  3. Sum of roots always negative.

  4. Product of roots always positive.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since $a,b,c$ are in A.P., so,

$2b = a + c$

$4{b^2} = {\left( {a + c} \right)^2}$

The discriminant of the given function$f\left( x \right) = 3a{x^2} - 4bx + c$ is,

$D = 16{b^2} - 12ac$

$ = 4{\left( {a + c} \right)^2} - 12ac$

$ = 4\left[ {\left( {{a^2} + {c^2} + 2ac} \right) - 3ac} \right]$

$ = 4\left( {{a^2} + {c^2} - ac} \right)$

$ = 4\left( {{a^2} + {c^2} - 2ac + ac} \right)$

$ = 4\left( {{{\left( {a - c} \right)}^2} + ac} \right)$

Case 1: If $a$ and$c$ are of opposite signs, then, $D = \left(  +  \right){\rm{ve}}$.

Case 2: If $a$ and$c$ are of same signs, then, $D = \left(  +  \right){\rm{ve}}$.

This shows that $f\left( x \right) = 0$ has two unequal real roots.