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Hint: We can assume the amount of first prize to be some positive value x. Now we can write the amount of second prize to be 20 less than x, third amount to be 20 less than the second prize and so on. Sum of all these will be equal to Rs. 700.

“Complete step-by-step answer:”

Let us assume the amount of first prize to be Rs. x. Now we can write the amount of other prizes to be following-

Second prize= $x-20$

Third prize=(Second prize)-20

\[\begin{array}{*{35}{l}}

=\left( x-20 \right)-20 \\

~=x-2\cdot 20 \\

\end{array}\]

Fourth prize=(Third prize)-20

$\begin{align}

& =(x-2\cdot 20)-20 \\

& =x-3\cdot 20 \\

\end{align}$

Likewise, we can write the fifth, sixth and seventh prize to be $(x-4\cdot 20)$ , $(x-5\cdot 20)$ and $(x-6\cdot 20)$ .

Now the sum of all of these prizes is Rs. 700. We can write this statement as an equation which is:

$x+(x-20)+(x-2\cdot 20)+(x-3\cdot 20)+(x-4\cdot 20)+(x-5\cdot 20)+(x-6\cdot 20)=700$

On simplifying we get,

$\begin{align}

& 7\cdot x-(1+2+3+4+5+6)\cdot 20=700 \\

& \Rightarrow 7x-21\times 20=700 \\

\end{align}$

Dividing both sides with 7 we have

$\begin{align}

& x-3\times 20=100 \\

& \Rightarrow x-60=100 \\

& \Rightarrow x=160 \\

\end{align}$

Therefore, the first prize is Rs. 160

Now we can calculate the other prizes as follows-

First prize= $Rs.160$

Second prize= $160-20=Rs.140$

Third prize= $140-20=Rs.120$

Fourth prize= $120-20=Rs.100$

Fifth prize= $100-20=Rs.80$

Sixth prize= $80-20=Rs.60$

Seventh prize= $60-20=Rs.40$

Note: For summation of $1+2+3+4+5+6$ instead of directly adding we could have used the formula $\sum\limits_{r=1}^{n}{r}=\dfrac{n(n+1)}{2}$ . Using this formula the summation $1+2+3+4+5+6$ can be written as $\sum\limits_{r=1}^{6}{r}=\dfrac{6\times 7}{2}=21$ which we can check that it is correct.

“Complete step-by-step answer:”

Let us assume the amount of first prize to be Rs. x. Now we can write the amount of other prizes to be following-

Second prize= $x-20$

Third prize=(Second prize)-20

\[\begin{array}{*{35}{l}}

=\left( x-20 \right)-20 \\

~=x-2\cdot 20 \\

\end{array}\]

Fourth prize=(Third prize)-20

$\begin{align}

& =(x-2\cdot 20)-20 \\

& =x-3\cdot 20 \\

\end{align}$

Likewise, we can write the fifth, sixth and seventh prize to be $(x-4\cdot 20)$ , $(x-5\cdot 20)$ and $(x-6\cdot 20)$ .

Now the sum of all of these prizes is Rs. 700. We can write this statement as an equation which is:

$x+(x-20)+(x-2\cdot 20)+(x-3\cdot 20)+(x-4\cdot 20)+(x-5\cdot 20)+(x-6\cdot 20)=700$

On simplifying we get,

$\begin{align}

& 7\cdot x-(1+2+3+4+5+6)\cdot 20=700 \\

& \Rightarrow 7x-21\times 20=700 \\

\end{align}$

Dividing both sides with 7 we have

$\begin{align}

& x-3\times 20=100 \\

& \Rightarrow x-60=100 \\

& \Rightarrow x=160 \\

\end{align}$

Therefore, the first prize is Rs. 160

Now we can calculate the other prizes as follows-

First prize= $Rs.160$

Second prize= $160-20=Rs.140$

Third prize= $140-20=Rs.120$

Fourth prize= $120-20=Rs.100$

Fifth prize= $100-20=Rs.80$

Sixth prize= $80-20=Rs.60$

Seventh prize= $60-20=Rs.40$

Note: For summation of $1+2+3+4+5+6$ instead of directly adding we could have used the formula $\sum\limits_{r=1}^{n}{r}=\dfrac{n(n+1)}{2}$ . Using this formula the summation $1+2+3+4+5+6$ can be written as $\sum\limits_{r=1}^{6}{r}=\dfrac{6\times 7}{2}=21$ which we can check that it is correct.

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