Let the first place team got ₹ $a$.

Since, award money increases by the same amount for successive finishing places.

Therefore series is an AP.

Let the constant amount be $d$.

$$\begin{aligned} &\begin{array}{ll} \text { Here, } & l=275, n=16 \text { and } S_{16}=8000 \\ \therefore & l=a+(n-) d \\ \Rightarrow & l=a+(16-1)(-d) \end{array}\\ &\text { [we take common difference ($-$ve) because series is decreasing] } \end{aligned}$$

$$\begin{array}{ll} \Rightarrow & 275=a-15 d \quad \text{... (i)}\\ \text { and } & S_{16}=\frac{16}{2}[2 a+(n-1) \cdot(-d)] \\ \Rightarrow & 8000=8[2 a+(16-1)(-d)] \\ \Rightarrow & 8000=8[2 a-15 d] \\ \Rightarrow & 1000=2 a-15 d \quad \text{... (ii)} \end{array}$$

On subtracting Eq. (i) from Eq. (ii), we get

$$\begin{aligned} & & (2 a-15 d)-(a-15 d) & =1000-275 \\ \Rightarrow & & 2 a-15 d-a+15 d & =725 \\ \therefore & & a & =725 \end{aligned}$$

Hence, first place team receive ₹ $725$.

Since, $a_1, a_2, a_3, \ldots, a_n$ are in AP.

$$\Rightarrow \quad a_2-a_1=a_3-a_2=\ldots=a_n-a_{n-1}=d \quad \text { [common difference] }$$

If $a_2-a_1=d$, then $\left(\sqrt{a_2}\right)^2-\left(\sqrt{a_1}\right)^2=d$

$$\begin{array}{ll} \Rightarrow & \left(\sqrt{a_2}-\sqrt{a_1}\right)\left(\sqrt{a_2}+\sqrt{a_1}\right)=d \\ \Rightarrow & \frac{1}{\sqrt{a_1}+\sqrt{a_2}}=\frac{\sqrt{a_2}-\sqrt{a_1}}{d} \end{array}$$

$$\begin{aligned} &\text { Similarly, }\\ &\begin{aligned} & \frac{1}{\sqrt{a_2}+\sqrt{a_3}}=\frac{\sqrt{a_3}-\sqrt{a_2}}{d} \\ & \qquad ... \qquad ... \qquad ...\\ & \qquad ... \qquad ... \qquad ...\\ & \qquad ... \qquad ... \qquad ...\\ & \frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}=\frac{\sqrt{a_n}-\sqrt{a_{n-1}}}{d} \end{aligned} \end{aligned}$$

On adding these terms, we get

$$\begin{aligned} & \frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}} \quad \text{[using above relations]}\\ = & \frac{1}{d}\left[\sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_2}+\ldots+\sqrt{a_n}-\sqrt{a_{n-1}}\right] \\ = & \frac{1}{d}\left[\sqrt{a_n}-\sqrt{a_1}\right] \quad \text{.... (i)} \end{aligned}$$

Again, $\quad a_n=a_1+(n-1) d \quad\left[\because T_n=a+(n-1) d\right]$

$$\begin{aligned} &\begin{array}{ll} \Rightarrow & a_n-a_1=(n-1) d \\ \Rightarrow & \left(\sqrt{a_n}\right)^2-\left(\sqrt{a_1}\right)^2=(n-1) d \\ \Rightarrow & \left(\sqrt{a_n}-\sqrt{a_1}\right)\left(\sqrt{a_n}+\sqrt{a_1}\right)=(n-1) d \Rightarrow \sqrt{a_n}-\sqrt{a_1}=\frac{(n-1) d}{\sqrt{a_n}+\sqrt{a_1}} \end{array}\\ &\text { On putting this value in Eq. (i), we get }\\ &\begin{gathered} \frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}} \\ =\frac{(n-1) d}{d\left(\sqrt{a_n}+\sqrt{a_1}\right)}=\frac{n-1}{\sqrt{a_n}+\sqrt{a_1}}\quad \text{Hence proved.} \end{gathered} \end{aligned}$$

Given series, $\left(3^3-2^3\right)+\left(5^3-4^3\right)+\left(7^3-6^3\right)+\quad \text{.... (i)}$

$$=\left(3^3+5^3+7^3+7\right)-\left(2^3+4^3+6^3+\ldots\right)$$

Let $T_n$ be the $n$th term of the series (i),

$$ \begin{aligned} \text { then }\quad T_n & =\left(n \text { thterm of } 3^3, 5^3, 7^3, \ldots\right)-\left(n \text { thterm of } 2^3, 4^3, 6^3, \ldots\right)=(2 n+1)^3-(2 n)^3 \\ & =(2 n+1-2 n)\left[(2 n+1)^2+(2 n+1) 2 n+(2 n)^2\right]\left[\because a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right] \\ & =\left[4 n^2+1+4 n+4 n^2+2 n+4 n^2\right]=\left[12 n^2+6 n\right]+1 \end{aligned}$$

(i) Let $S_n$ denote the sum of $n$ term of series (i).

Then,

$$\begin{aligned} S_n & =\Sigma T_n=\Sigma\left(12 n^2+6 n\right) \\ & =12 \Sigma n^2+6 \Sigma n+\Sigma n \\ & =12 \cdot \frac{n(n+1)(2 n+1)}{6}+\frac{6 n(n+1)}{2}+n \\ & =2 n(n+1)(2 n+1)+3 n(n+1)+n \\ & =2 n(n+1)(2 n+1)+3 n(n+1)+n \\ & =\left(2 n^2+2 n\right)(2 n+1)+3 n^2+3 n+n \\ & =4 n^3+2 n^2+4 n^2+2 n+3 n^2+3 n+n \\ & =4 n^3+9 n^2+6 n \end{aligned}$$

(ii) Sum of 10 terms,

$$\begin{aligned} S_{10} & =4 \times(10)^3+9 \times(10)^2+6 \times 10 \\ & =4 \times 1000+9 \times 100+60 \\ & =4000+900+60=4960 \end{aligned}$$

Given that, sum of $n$ terms of an AP,

$$\begin{aligned} S_n & =2 n+3 n^2 \\ T_n & =S_n-S_{n-1} \\ & =\left(2 n+3 n^2\right)-\left[2(n-1)+3(n-1)^2\right] \\ & =\left(2 n+3 n^2\right)-\left[2 n-2+3\left(n^2+1-2 n\right)\right] \\ & =\left(2 n+3 n^2\right)-\left(2 n-2+3 n^2+3-6 n\right) \\ & =2 n+3 n^2-2 n+2-3 n^2-3+6 n \\ & =6 n-1 \\ \therefore \quad \text { rth term } T_r & =6 r-1 \end{aligned}$$

Let the numbers be $a$ and $b$.

Then,

$$\begin{aligned} A & =\frac{a+b}{2} \\ \Rightarrow \quad 2 A & =a+b\quad \text{.... (i)} \end{aligned}$$

and $G_1, G_2$ be geometric mean between $a$ and $b$, then $a, G_1, G_2, b$ are in GP.

Let $r$ be the common ratio.

Then,

$$\begin{aligned} & b=a r^{4-1} \quad \left[\because a_n=a r^{n-1}\right] \\ \Rightarrow\quad& b=a r^3 \Rightarrow \frac{b}{a}=r^3 \\ \therefore \quad& r=\left(\frac{b}{a}\right)^{1 / 3} \end{aligned}$$

Now, $$G_1=a r=a\left(\frac{b}{a}\right)^{1 / 3} \quad\left[\because r=\left(\frac{b}{a}\right)^{1 / 3}\right]$$

and $$G_2=a r^2=a\left(\frac{b}{a}\right)^{2 / 3}$$

$$\begin{aligned} \text{RHS} & =\frac{G_1^2}{G_2}+\frac{G_2^2}{G_1}=\frac{\left[a\left(\frac{b}{a}\right)^{1 / 3}\right]^2}{a\left(\frac{b}{a}\right)^{2 / 3}}+\frac{\left[a\left(\frac{b}{a}\right)^{2 / 3}\right]^2}{a\left(\frac{b}{a}\right)^{1 / 3}} \\ & =\frac{a^2\left(\frac{b}{a}\right)^{2 / 3}}{a\left(\frac{b}{a}\right)^{2 / 3}}+\frac{a^2\left(\frac{b}{a}\right)^{4 / 3}}{a\left(\frac{b}{a}\right)^{1 / 3}} \\ & =a+a\left(\frac{b}{a}\right)=a+b=2 A \quad \text{[using Eq. (i)]}\\ & =\text { LHS } \end{aligned}$$