Complete the a.p series with given information - class-XI

Given series $2,a,8$

The equation cos(px) + cos(qx) = 0 implies 2cos((p+q)x/2)cos((p-q)x/2) = 0. Solving for x gives the roots in AP with the specified common differences.

The sequence is 3, 7, 14, 24. The differences are 4, 7, 10. This is a quadratic sequence. Testing n=1 gives 3, n=2 gives 7. Option B fits the pattern.

By calculating the distances from A to the lines, one can show that the segments form an arithmetic progression based on the properties of the lines and the point A.

$2+3+6+11+18+.................$

$a, b, c \in A.P. \quad \left( \text{Given} \right)$

Lets find the difference between the numbers of the series:

Let a and d be the first term and common difference of an AP.
Since, $\displaystyle { T } _{ m }=\frac { 1 }{ n } $
$\displaystyle \therefore a+\left( m-1 \right) d=\frac { 1 }{ n } $....(i)
and $\displaystyle { T } _{ n }=\frac { 1 }{ m } $
$\displaystyle \Rightarrow a+\left( n-1 \right) d=\frac { 1 }{ m } $.....(ii)
On solving Eqs. (i) and (ii), we get
$\displaystyle a=\frac { 1 }{ mn } and\quad d=\frac { 1 }{ mn } $
$\displaystyle \therefore \quad { T } _{ mn }=a+\left( mn-1 \right) d$

If you will subtract or add a constant number to each terms of an AP, then later sequence will also be in AP with same common difference

Let $a$ be first term $d$ be common difference of an AP having $n$ number of terms.

$\dfrac{1}{x},\dfrac{2}{x},\dfrac{3}{x},....$ is a property of A.P.
Here the constant x is divided from each term of an A.P. with same common difference.

Let $a$ be first term $d$ be common difference of an AP having $n$ number of terms.

Required numbers are $24,30,36,40,.....96$
This is an A.P. in which $a=24,d=6,l=96$
Let the number of terms in it be $n$.
Then ${t} _{n}=96$ $\Rightarrow$ $a+(n-1)d=96$
$\Rightarrow$ $24+(n-1)\times 6=96$
$\Rightarrow$ $(n-1)=12$
$\Rightarrow$ $n=13$
Required number of numbers $=13$

Let the first term be $a$ and common difference be $d$.
So, sum of first $10$ terms $=\dfrac { 10 }{ 2 } (2a+(10-1)d)$

$T _m+ _1-T _m=15$ 

$\Rightarrow ^{(m+1)}C _3-^{(m)}C _3=15$

By verification $m=6$ by solve

$20, \displaystyle 19\frac{1}{4}, 18\frac{1}{2}, ....$
or $\displaystyle 20, \frac{77}{4}, \frac{37}{2}, ....$
$a=20$
$d=\displaystyle\frac{77}{4}-20=\frac{-3}{4}$
Let $n^{th}$ term of A.P. be first negative term
So, $20+(n-1)\left(\displaystyle\frac{-3}{4}\right)<0$
$\Rightarrow 80-3n+3<0$
$\Rightarrow 3n>83$
$\Rightarrow n > 27\displaystyle\frac{2}{3}$
Hence, $28^{th}$ term is first negative term.
(Option $3$).

T_m = mC3. The condition T_{m+1} - T_m = 15 becomes (m+1)C3 - mC3 = 15, which simplifies to mC2 = 15. Solving m(m-1)/2 = 15 gives m^2 - m - 30 = 0, so (m-6)(m+5)=0. Thus m=6.

Clearly $x=1$ is a solution
$\therefore$  product of the roots $=\dfrac { a-b }{ b-c }$ 
$\therefore \left( 1 \right) \left( 1 \right) =\dfrac { a-b }{ b-c }$ 
$\Longrightarrow b-c=a-b$
$\Longrightarrow2b=a+c\Longrightarrow a,b,c$ are in Arithmetic progression.

Consider mth term of an AP 
$t _m=a+(m-1)d$
Now, consider (n-m)th term:
$t _{n-m}=a+(n-m-1)d$
Sum $= 2a+(n-1)d $

$A,B, C$ in $AP$

The $three$ middle most terms of $15$ terms is $69.$

$x = A{R^{p - 1}}$

Using the identity sin(A)/cos(B) related to telescoping sums, the expression simplifies. For f_2, the calculation results in -1.

$a,b,c$ are in A.P 

Given 

We have,

If A, B, C are in AP, then 2B = A+C. Since A+B+C = 180, B=60. Using the cosine rule b^2 = a^2 + c^2 - 2ac cos(B), and cos(60)=1/2, we get b^2 = a^2 + c^2 - ac.

if $a,b,c$ is in $AP$ then we know that,

Let the four parts be $a-3d,,a-d,,a+d$ and $a+3d$

$Mean=\dfrac{S}{n}$

$\log { 2 } ,\log { \left( { 2 }^{ x }-1 \right)  } ,\log { \left( { 2 }^{ x }-3 \right)  } $ are in AP

Given 1/a, 1/b, 1/c are in AP, 2/b = 1/a + 1/c. Substituting this into the expression and simplifying leads to the result 4/ac - 3/b^2.

The ex-radii are r1 = Area/(s-a), r2 = Area/(s-b), r3 = Area/(s-c). If these are in HP, then (s-a), (s-b), (s-c) are in AP, which implies -a, -b, -c are in AP, so a, b, c are in AP.

Let $a,A$ be first term and $d,D$ be common difference and $S _n,s _n$ be there respective sum.

If a_n and b_n are APs, their sum a_n + b_n is also an AP. The common difference of the sum is the sum of the individual common differences.

Given roots of the polynomial are in AP

Given $x,yz$ are in $AP$

We have,

Given equation is

$\left( a-b \right){{x}^{2}}+\left( c-a \right)x+\left( b-c \right)=0$

On comparing that,

$A{{x}^{2}}+Bx+C=0$

Now,

$ A=\left( a-b \right) $

$ B=\left( c-a \right) $

$ C=\left( b-c \right) $

Roots are equal

Then,

$ D=0 $

$ {{B}^{2}}-4AC=0 $

$ \Rightarrow {{\left( c-a \right)}^{2}}-4\left( a-b \right)\left( b-c \right)=0 $

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

$ \Rightarrow {{c}^{2}}+{{a}^{2}}-2ac=4ab-4ac-4{{b}^{2}}+4bc $

$ \Rightarrow {{c}^{2}}+{{a}^{2}}-2ac+4ac=4ab-4{{b}^{2}}+4bc $

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

Hence, this is the answer

$a,b,c$ are in AP
$\Rightarrow (abc+1)a, (abc+1)b,(abc+1)c $ are in AP
$\Rightarrow  \dfrac{(abc+1)a}{abc},\dfrac{(abc+1)b}{abc},\dfrac{(abc+1)c}{abc}$ are in AP
$\Rightarrow a+\dfrac{1}{bc},b+\dfrac{1}{ac},c+\dfrac{1}{ab} $ are in AP
$\therefore$ Ans. is option A.

${ a } _{ 10 }={ a } _{ 1 }+9{ d } _{ 1 }\qquad \Longrightarrow { d } _{ 1 }=\cfrac { 3-2 }{ 9 } =\cfrac { 1 }{ 9 } $

The given sequence is an A.P. with first term $a=72$ and common difference $d=-2$. Let its nth term be 40.

Let $d$ be the common difference of the $A.P.$

The given series is $4,13,22,31,40,....$

T_n = a + (n-1)d. T_3 = a + 2d = 8 and T_7 = a + 6d = 24. Subtracting gives 4d = 16, so d = 4. Then a + 8 = 8, so a = 0. T_n = 0 + (n-1)4 = 4n - 4. If the question asks for T_1, it is 0; if it asks for T_0 (not standard), it is -4.

First term of $AP = P$

Consecutive terms of $AP \Rightarrow 2x, (x + 10 ) , (3x + 2)$

Given,

Number of integers between 2$^n$ and $2^{2n+ 1} - 2^n - 11$
and I term $= 2^n + 1$
last term $= 2^{n + 1} - 1$
$\therefore \displaystyle S _n = \frac{(2^{n + 1} - 2^n - 1)}{2} [2^n + 1 + 2^{n + 1} - 1]$
$\displaystyle \frac{(2^{n + 1} - 2^n - 1)}{2} (2^n)(1 + 2)$
$= (2^n - 1) \displaystyle \frac{(2^n).3}{2}$
$S _n= 9 \lambda; \lambda \varepsilon I$
$\therefore 3 \times 2^{n - 1} \times (2^n - 1) = 9 \lambda$
$2^{n - 1} \times (2^n - 1) = 3 \lambda$
$2^n (2^n - 1) = 6 \lambda$
It is possible when n is even.

Let the numbers are be $a - d, a, a + d$
Then $a - d + a + a + d = 21$
$3a = 21$
$a = 7$
and $(a - d)(a + d) = 45$
$a^2 - d^2 = 45$
$d^2 = 4$
$d=\pm 2$
Hence, the numbers are $5, 7$ and $9$ when $d = 2$ and $9, 7$ and $5$ when $d = -2$. In both the cases numbers are the same.

Let the first term be $a$ and common difference be $d$.
So, sum of first $10$ terms $=\dfrac { 10 }{ 2 } (2a+(10-1)d)$

We know that the sum of terms of AP equidistant from the beginning and end is always same and it is always equal to the sum of first and last terms.
$\Rightarrow { a } _{ 1 }+{ a } _{ 24 }={ a } _{ 6 }+{ a } _{ 20 }={ a } _{ 10 }+{ a } _{ 15 }$
$\because { a } _{ 1 }+{ a } _{ 5 }+{ a } _{ 10 }+{ a } _{ 15 }+{ a } _{ 20 }+{ a } _{ 24 }=225$
$\therefore 3\left( { a } _{ 1 }+{ a } _{ 24 } \right) =225\Rightarrow { a } _{ 1 }+{ a } _{ 24 }=75$
$\therefore { S } _{ 24 }=\dfrac { 24 }{ 2 } \left( { a } _{ 1 }+{ a } _{ 24 } \right)$   $\left[ \because { S } _{ n }=\dfrac { n }{ 2 } \left( { a } _{ 1 }+{ a } _{ n } \right)  \right] $
           $=12\left( 75 \right) =900=9\times { 10 }^{ 2 }$

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

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.

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

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.

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

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

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.

$first, , term, , a=-3 $

 $\displaystyle \frac{b+c-a}{a},\frac{c+a-b}{b},\frac{a+b-c}{c}$ are in $AP$

Given: