A sum of ` 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ` 20 less than its preceding prize, find the value of each of the prizes.
Step-by-Step Solution
Key Concept: The cash prizes form an arithmetic progression (AP) with first term \(a\) and common difference \(d = -20\). Use the sum formula for an AP: \(S_n = \frac{n}{2}[2a + (n-1)d]\) and solve for \(a\). Then generate the remaining terms using \(a_k = a + (k-1)d\).
1. Identify the AP parameters
- Number of prizes (terms) : \(n = 7\)
- Common difference (each prize is \(20\) less than the previous one) : \(d = -20\)
- Let the first prize be \(a\) rupees.
2. Write the sum of the AP
\[ S_n = \frac{n}{2}\bigl[2a + (n-1)d\bigr] \]
Given \(S_n = 700\), substitute \(n = 7\) and \(d = -20\):
\[ 700 = \frac{7}{2}\bigl[2a + 6(-20)\bigr] \]
3. Simplify the equation
\[ 700 = \frac{7}{2}\bigl[2a - 120\bigr] \]
Multiply both sides by 2:
\[ 1400 = 7\bigl[2a - 120\bigr] \]
Divide by 7:
\[ 200 = 2a - 120 \]
Add 120 to both sides:
\[ 320 = 2a \]
Hence \(a = \frac{320}{2} = 160\).
4. Find each prize using \(a_k = a + (k-1)d\):
- 1st prize: \(a_1 = 160\) rupees
- 2nd prize: \(a_2 = 160 + (1)(-20) = 140\) rupees
- 3rd prize: \(a_3 = 160 + (2)(-20) = 120\) rupees
- 4th prize: \(a_4 = 160 + (3)(-20) = 100\) rupees
- 5th prize: \(a_5 = 160 + (4)(-20) = 80\) rupees
- 6th prize: \(a_6 = 160 + (5)(-20) = 60\) rupees
- 7th prize: \(a_7 = 160 + (6)(-20) = 40\) rupees
5. Verification
\[ 160 + 140 + 120 + 100 + 80 + 60 + 40 = 700 \]
The total matches the given sum, confirming the solution.
Therefore, the cash prizes are Rs. 160, 140, 120, 100, 80, 60 and 40 respectively.
Correct Answer: First prize = Rs.160; subsequent prizes = Rs.140, Rs.120, Rs.100, Rs.80, Rs.60, Rs.40.