Tag: problems on ap

Questions Related to 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 $x _{3} + x _{8} + x _{22} = ?$

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

  1. $-21$

  2. $-15$

  3. $-18$

  4. $-31$

Show answer
Correct Option: A
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, $x _{3}+x _{8}+x _{22}=3a+30d = -21$

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

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

  1. $n^2-1$

  2. $(2n-1)^2$

  3. $n^2$

  4. $n^2+1$

Show answer
Correct Option: C
Explanation:
$a _n=(2n-1)$
$\Rightarrow$  $a _1=2\times 1-1=1$
$\Rightarrow$  $a _2=2\times 2-1$
            $=4-1$
            $=3$
$\Rightarrow$  $d=a _1-a _1=3-1$
$\therefore$  $d=2$
$\Rightarrow$  $S _1=\dfrac{n}{2}[2a _1+(n-1)d]$
           
            $=\dfrac{n}{2}[2(1)+(n-1)2]$

            $=\dfrac{n}{2}[2+2n-2]$

            $=\dfrac{n}{2}\times 2n$

            $=n^2$

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

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

  1. Arithmetic progression

  2. Geometric prograsson

  3. Harmonic prograssiion

  4. Arithmetco- Geometric prograssion

Show answer
Correct Option: A

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 _________.

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

  1. $0$

  2. $1$

  3. pqr

  4. p$+$qr

Show answer
Correct Option: A
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

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-

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

  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

Show answer
Correct Option: A
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.

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

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

  1. $123968$

  2. $133068$

  3. $133268$

  4. $187332$

Show answer
Correct Option: C
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.$

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
$x+y\ \tan { \left( \dfrac { \alpha +\beta  }{ 2 }  \right)  } =0$

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

  1. $-1$

  2. $0$

  3. $\sin (\alpha + \beta /2)$

  4. $\sin (\alpha - \beta /2)$

Show answer
Correct Option: D
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

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

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

  1. 525

  2. 563

  3. 184

  4. 189

Show answer
Correct Option: A
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 
$S _{n}=525$

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

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

  1. an odd number

  2. an even number

  3. a prime number

  4. a composite number

Show answer
Correct Option: D
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.

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

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

  1. No real solution.

  2. Two unequal real roots.

  3. Sum of roots always negative.

  4. Product of roots always positive.

Show answer
Correct Option: B
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.