A sum of 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 Rs 20 less than its preceding prize, find the value of each of the prizes.
The prize amounts are 160, 140, 120, 100, 80, 60, 40.
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.