Ex.5.3 Q16 Arithmetic Progressions Solution - NCERT Maths Class 10

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A sum of \(\rm{Rs}\, 700\) is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is \(\rm{Rs}\,20\) less than its preceding prize, find the value of each of the prizes.

 Video Solution
Arithmetic Progressions
Ex 5.3 | Question 16

Text Solution

What is Known?

\(7\) cash prizes are given, and each prize is \(\rm{Rs}\,20\) less than its preceding prize.

What is Unknown?

Value of each of the prizes


General form of an arithmetic progression is \(a,\left( {a + d} \right),\left( {a + 2d} \right),\left( {a + 3d} \right),\dots\)

Sum of the first \(n\)terms of an AP is given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\)

Where \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms.


Let the cost of 1st prize be \(x\).

Then the cost of 2nd prize \(=x-20\)

And the cost of 3rd prize \( = x - 40\)

Prizes are \(x,{\rm{ }}\left( {x - 20} \right),{\rm{ }}\left( {x - 40} \right),\dots\)

By observation that the costs of these prizes are in an A.P., having common difference as \(−20\) and first term as \(x\).

\[\begin{align} a&=x \\ d& =-20 \\\end{align}\]

Given that, \({S_7} = 700\)

\({S_n}= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\)

\[\begin{align}\frac{7}{2}\left[ {2x + \left( {7 - 1} \right)d} \right] &= 700\\\left[ {2x + \left( 6 \right) \times \left( { - 20} \right)} \right] &= 200\\x + 3 \times \left( { - 20} \right) &= 100\\x - 60 &= 100\\x &= 160\end{align}\]

Therefore, the value of each of the prizes was \(\rm{Rs}\,160,\) \(\rm{Rs} \,140,\) \(\rm{Rs}\, 120, \;\rm{Rs}\,100, \;\rm{Rs}\,80, \;\rm{Rs}\,60,\) and \(\rm{Rs} \,40.\)