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**Hint:**We will have to make a linear equation taking the last prize as an unknown value and relate the remaining prizes to make an equation. Here, each prize is some amount less than its preceding prize. So clearly, it is a problem of arithmetic series. Firstly, the amount of the lowest prize should be assumed & others are to be supposed accordingly.

**Complete step-by-step answer:**

Let as assume the value of the lowest prize be \[Rs.{\text{ }}x\] .

Total sum of cash prizes = Rs. 980 is to be divided into $7$ students.

Now according to given information in the question, since each prize is \[Rs.{\text{ }}30\] less than its preceding prize, \[Rs.{\text{ }}30\] can be added to each prize of its succeeding prize,

So, here we have to apply concepts of arithmetic series because there is certain common difference between amount of two consecutive prizes:

Adding the amount of each of 7 prizes & equalizing it to $Rs.980$

$\therefore x + \left( {x + 30} \right) + \left( {x + 2 \times 30} \right) + ....$ $ + \left( {x + 6 \times 30} \right) = 980$

$ \Rightarrow 7x + 30\left( {1 + 2 + 3 + 4 + 5 + 6} \right) = 980$

[ adding $x$ term from each of $7$ terms & taking $30$ as common multiple]

$ \Rightarrow 7x + 30 \times \dfrac{{6 \times \left( {6 + 1} \right)}}{2} = 980$ $\left[ {\because 1 + 2 + 3 + ..... + n = \dfrac{{n\left( {n + 1} \right)}}{2}} \right]$

$ \Rightarrow 7x + 630 = 980$ [ simplifying above equation ]

$ \Rightarrow x = \dfrac{{980 - 630}}{7} = \dfrac{{350}}{7} = 50$

$\therefore $ As considered , putting the value of x, amount of each prizes is –

$50,\left( {50 + 30} \right),\left( {50 + 2 \times 30} \right),....$

The prizes are \[Rs.{\text{ }}50,{\text{ }}Rs{\text{ }}80,{\text{ }}Rs{\text{ }}110,{\text{ }}Rs{\text{ }}140,{\text{ }}Rs{\text{ }}170,{\text{ }}Rs{\text{ }}210,{\text{ }}Rs{\text{ }}240.\].

**Note:**In this type of problem of arithmetic series, the solution provided is done considering each prize of the students as succeeding, so that addition can be used. This can also be solved by subtraction of the highest value and then its preceding prize. Do the calculation carefully to avoid mistakes. Hence the concept of arithmetic series should be clear. Key formula used here is $\left[ {\because 1 + 2 + 3 + ..... + n = \dfrac{{n\left( {n + 1} \right)}}{2}} \right]$

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