Tag: complete the a.p series with given information
Questions Related to complete the a.p series with given information
$x _{1}, x _{2}, x _{3}, ....$ are in A.P.
If $x _{1} + x _{7} + x _{10} = -6$ and $x _{3} + x _{8} + x _{12} = -11$, then $x _{3} + x _{8} + x _{22} = ?$
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$-21$
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$-15$
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$-18$
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$-31$
Let the first term be $a$ and common difference be $d$
If the $n^{th}$ term of an AP be $(2n-1)$, then the sum of its first n terms will be.
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$n^2-1$
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$(2n-1)^2$
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$n^2$
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$n^2+1$
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
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Arithmetic progression
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Geometric prograsson
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Harmonic prograssiion
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Arithmetco- Geometric prograssion
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 _________.
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$0$
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$1$
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pqr
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p$+$qr
let $a$ be first term of A.P. and $ d$ is it's common difference
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-
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$\left( \dfrac { -\pi }{ 2 } ,\dfrac { \pi }{ 2 } \right)$
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$\left( \dfrac { -\pi }{ 3 } ,\dfrac { \pi }{ 3 } \right)$
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$\left( \dfrac { -\pi }{ 6 } ,\dfrac { \pi }{ 6 } \right)$
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none of these
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?
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$123968$
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$133068$
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$133268$
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$187332$
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
$x+y\ \tan { \left( \dfrac { \alpha +\beta }{ 2 } \right) } =0$
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$-1$
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$0$
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$\sin (\alpha + \beta /2)$
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$\sin (\alpha - \beta /2)$
Find the sum of the first $15$ terms of the following sequences having $n$th term as
${a} _{n}=3+4n$
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525
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563
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184
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189
$a _{n}=3+4n$ (Given)
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
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an odd number
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an even number
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a prime number
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a composite number
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
$\Rightarrow{T} _{r}=\left(r+1-r\right)\left({\left(r+1\right)}^{2}+r\left(r+1\right)+{r}^{2}\right)+\frac{1}{2}-2$
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
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No real solution.
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Two unequal real roots.
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Sum of roots always negative.
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Product of roots always positive.
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.