Understanding the Problem
In an arithmetic progression (AP), the nth term can be expressed as:
- \( a_n = a + (n-1)d \)
Where:
- \( a \) = first term
- \( d \) = common difference
We need to show that the 72nd term is 4 times the 15th term given that the 24th term is twice the 10th term.
Step 1: Establish the Given Condition
We know that:
- \( a_{24} = 2 \times a_{10} \)
Using the formula for the terms of the AP:
- \( a + 23d = 2(a + 9d) \)
Simplifying this equation:
- \( a + 23d = 2a + 18d \)
Rearranging gives:
- \( a + 23d - 2a - 18d = 0 \)
Thus:
- \( -a + 5d = 0 \)
- Therefore, \( a = 5d \)
Step 2: Find the 72nd and 15th Terms
Now, we need to find the 72nd term \( a_{72} \) and the 15th term \( a_{15} \):
- \( a_{72} = a + 71d \)
- \( a_{15} = a + 14d \)
Substituting \( a = 5d \):
- \( a_{72} = 5d + 71d = 76d \)
- \( a_{15} = 5d + 14d = 19d \)
Step 3: Show the Relationship
Now, we need to check if \( a_{72} = 4 \times a_{15} \):
- \( 4 \times a_{15} = 4 \times 19d = 76d \)
Since both \( a_{72} \) and \( 4 \times a_{15} \) equal \( 76d \), we conclude that:
- \( a_{72} = 4 \times a_{15} \)
Conclusion
Thus, it is proven that the 72nd term of the AP is indeed four times the 15th term, satisfying the condition outlined.