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Let a, ar, ar², . an increasing G.P then r > 1 & a > 0

given : ar + ar5 = $\frac{25}{2}$ .(i)

and $a{r}^{2}a{r}^4=25\Rightarrow a^2{r}^6=25\Rightarrow {\left(a{r}^3\right)}^2=25$

$\Rightarrow a{r}^3=5.......\left(ii\right)$

From (i) and (ii), $\frac{ar\left(1+r^4\right)}{a{r}^3}=\frac{25}{2\times 5}$

$\Rightarrow 2+2r^4=5r^2\Rightarrow 2r^4-5r^2+2=0$

$\Rightarrow \left(2r^2-1\right)\left(r^2-2\right)=0$

$\Rightarrow r^2=\frac{1}{2},2\therefore r^2=2\Rightarrow r=\sqrt[]{2}$

$\therefore t_4+t_6+t_8=5+10+20=35$

R = { (P, Q) |P and Q are at the same distance from the origin}.

at $\left(1,-1\right),x^2+y^2={\left(1\right)}^2+{\left(-1\right)}^2=1+1=2$

Former US President, Barack Obama graduated with a BA in Political Science from the Columbia College, in 1983, and later enrolled in the Harvard Law School, becoming the first black president of the Harvard Law Review.  

After his studies at Harvard Law, Barack moved to Chicago, where he helped rebuild devastated communities, due to the closure of the local steel plants. He credits his experience there, in helping him find his true identity, that ultimately shaped his path to becoming the first black US president. 

$P_1:3x+15+21z=9$           

P₂ : x – 3y – z = 5

P₃ : 2x + 10y + 14z = 5

Ratio of the direction cosines of P₁ and P₂

$\frac{3}{2}=\frac{15}{10}=\frac{21}{14}$

Hence, P₁ and P₃ are parallel.

 $\overrightarrow{a}\times \left(\overrightarrow{a}\times \overrightarrow{b}\right)=\left(\overrightarrow{a}.\overrightarrow{b}\right)\overrightarrow{a}-{\left|\overrightarrow{a}\right|}^2\overrightarrow{b}=0-{\left|\overrightarrow{a}\right|}^2\overrightarrow{b}\left(as\overrightarrow{a}.\overrightarrow{b}=0given\right)$

$\overrightarrow{a}\times \left(\overrightarrow{a}\times \left(\overrightarrow{a}\times \overrightarrow{b}\right)\right)=-{\left|\overrightarrow{a}\right|}^2\overrightarrow{a}\times \overrightarrow{b}$

$\overrightarrow{a}\times \left(\overrightarrow{a}\times \left(\overrightarrow{a}\times \left(\overrightarrow{a}\times \overrightarrow{b}\right)\right)\right)=-{\left|\overrightarrow{a}\right|}^2\overrightarrow{a}\times \left(\overrightarrow{a}\times \overrightarrow{b}\right)=-{\left|\overrightarrow{a}\right|}^2\left(-{\left|\overrightarrow{a}\right|}^2\overrightarrow{b}\right)={\left|\overrightarrow{a}\right|}^4\overrightarrow{b}$

Admission to Sewa Community College, Andheri, is easy since the institute admits students based on merit. The institute offers UG, PG, Certificate and Diploma courses. The institute offers more than 12 courses in the stream of Healthcare. Candidates must meet the eligibility criteria set by the college. 

Yes, admissions for 2025 are open at Sewa Community College. SCC, Andheri admissions are currently open for all courses. Candidates are selected for courses based on merit. Aspirants can apply online. Interested candidates meeting the eligibility criteria can fill out the form. 

$T_{r+1}{=}^{10}{C}_r{\left(t{x}^{\frac{1}{5}}\right)}^{10-r}{\left(\frac{{\left(1-x\right)}^{\frac{1}{10}}}{t}\right)}^r$

According to question, 10 – 2r = 0 ⇒ r = 5

$\therefore T_6{=}^{10}{C}_5\times {\left(1-x\right)}^{\frac{1}{2}}$

T₆ is maximum, when f(x) = $x{\left(1-x\right)}^{\frac{1}{2}}$ is maximum.

$f\text{'}\left(x\right)={\left(1-x\right)}^{\frac{1}{2}}-\frac{x}{2\sqrt[]{1-x}}=\frac{2\left(1-x\right)-x}{2\sqrt[]{1-x}}$

For maximum, $f\text{'}\left(x\right)=0\Rightarrow x=\frac{2}{3}$

$\therefore T_6{=}^{10}{C}_5\frac{2}{3}.{\left(\frac{1}{3}\right)}^{\frac{1}{2}}=\frac{2\left(10!\right)}{3\sqrt[]{3}{\left(5!\right)}^2}$

Bollywood Actress Sara Ali Khan, graduated with a BA in History and Political Science from the Columbia University, before embarking on her acting journey in 2018. 

Her very first film, Kedarnath, bough her a Filmfare Award, a Screen Award and an IIFA Award. Some of her other notable works include: Zara Hatke Zara Bachke, Sky Force, Metro In Dino, Atrangi Re and Simmba. 

Sewa Community College admits students based on merit. The institute offers BVoc and MVoc courses. Apart from the UG and PG courses, the institute also offers Certificate and Diploma courses. Candidates must meet the eligibility criteria set by the college. To get admission at Sewa Community College, candidates must meet the eligibility criteria set by the college. For UG courses, aspirants must complete Class 12 and graduation for PG courses. 

Let n be the number of times.

$p=\frac{1}{2},q=\frac{1}{2}$

According to question,

${\Rightarrow }^n{C}_7{=}^n{C}_9\Rightarrow n-7=9\Rightarrow n=16$

$p\left(x=2\right){=}^{16}{C}_2{\left(\frac{1}{2}\right)}^2.{\left(\frac{1}{2}\right)}^{14}=\frac{15}{2^{13}}$

Put x = $\frac{1}{3}$

S = 1 + 2x + 7x² + 12x³ + 17x⁴ + 22x⁵ + .

$\underset{\_}{xS=x+2x^2+7x^3+12x^4+17x^5+.........}$

Subtracting,

$\left(1-x\right)S=1+x+5x^2+5x^3+5x^4+5x^5+......$

= $1-4x+5x+5x^2+5x^3+.........$

$=\frac{1-x-4x+4x^2+5x}{1-x}=\frac{4x^2+1}{1-x}$

$S=\frac{4x^2+1}{{\left(1-x\right)}^2}putx=\frac{1}{3}wegets=\frac{\frac{4}{9}+1}{\frac{4}{9}}=\frac{13}{4}$

Sewa Community College admissions are open online. Since the admission is open, candidates can apply. The institute offers BVoc and MVoc courses. Apart from the UG and PG courses, the institute also offers Certificate and Diploma courses. Candidates must complete Class 12 as the basic eligibility criteria for UG courses. 

 $\frac{dv}{dt}\propto V\Rightarrow \frac{dv}{dt}=\lambda V\Rightarrow {\displaystyle \underset{1000}{\overset{1200}{\int }}\frac{dv}{v}}=\lambda {\displaystyle \underset{0}{\overset{2}{\int }}dt}\Rightarrow \lambda =\frac{1}{2}\mathcal{l}n\frac{6}{5}$

${\displaystyle \underset{1000}{\overset{2000}{\int }}\frac{dv}{v}}=\frac{1}{2}\mathcal{l}n\left(\frac{6}{5}\right){\displaystyle \underset{0}{\overset{T}{\int }}dt}\Rightarrow T=\frac{2\mathcal{l}n2}{\mathcal{l}n\left(\frac{6}{5}\right)}\Rightarrow k=2ln2$