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

## Question

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.

## 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

**Reasoning:**

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.

**Steps:**

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 \(\rm{Rs} \,40.\)