NCERT Solutions
Class 10 - Mathematics - Chapter 5: ARITHMETIC PROGRESSIONS - Exercise 5.3

Question 16

Step 1: Recognise the AP pattern.

We have 7 prizes and each prize is Rs 20 less than the previous one. So the prizes form a decreasing arithmetic progression (AP).

Let the first (highest) prize be \(a\) rupees. Then the common difference is:

\[d = -20\]

Step 2: Write the 7 prizes as AP terms.

The 7 prizes (terms of the AP) will be:

\[a,\ a - 20,\ a - 40,\ a - 60,\ a - 80,\ a - 100,\ a - 120\]

Step 3: Use the sum condition.

Total money to be distributed is Rs 700. So, the sum of these 7 terms is 700.

Sum of first \(n\) terms of an AP is:

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

Here, \(n = 7\), \(d = -20\), \(S_7 = 700\). Substitute:

\[700 = \dfrac{7}{2}[2a + (7 - 1)(-20)]\]

\[700 = \dfrac{7}{2}[2a + 6(-20)]\]

\[700 = \dfrac{7}{2}[2a - 120]\]

Step 4: Solve for \(a\).

Multiply both sides by 2:

\[1400 = 7(2a - 120)\]

Divide by 7:

\[2a - 120 = 200\]

Add 120 to both sides:

\[2a = 320\]

\[a = 160\]

So, the highest prize is Rs 160.

Step 5: List all prize amounts.

Now generate the remaining six prizes by subtracting 20 each time:

1st prize: \(160\)
2nd prize: \(160 - 20 = 140\)
3rd prize: \(140 - 20 = 120\)
4th prize: \(120 - 20 = 100\)
5th prize: \(100 - 20 = 80\)
6th prize: \(80 - 20 = 60\)
7th prize: \(60 - 20 = 40\)

Step 6: Quick check of the sum.

Add them: \(160 + 140 + 120 + 100 + 80 + 60 + 40 = 700\). The total matches the given sum.

Conclusion: The seven prizes are Rs 160, 140, 120, 100, 80, 60, 40.