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A1, a2, a3 represents 1st, 2nd & 3rd terms of an AP
respectively, 1st term is 2 & (a1 + a2)a3 is minimum, then the
common difference is equal to
y,
y2,
y3 ………; if y > 2
ퟑ퐳
4y
3y?
respectively, 1st term is 2 & (a1 + a2)a3 is minimum, then the
common difference is equal to
y,
y2,
y3 ………; if y > 2
ퟑ퐳
4y
3y?
Most Upvoted Answer
A1, a2, a3 represents 1st, 2nd & 3rd terms of an AP respectively, 1st ...
Understanding the Problem
To solve the problem, we need to analyze the arithmetic progression (AP) where the first term \( a_1 = 2 \). The terms can be expressed as:
- \( a_1 = 2 \)
- \( a_2 = 2 + d \)
- \( a_3 = 2 + 2d \)
Here, \( d \) represents the common difference.
Expression for Minimization
We need to minimize the expression \( (a_1 + a_2) a_3 \):
- Calculate \( a_1 + a_2 \):
\( a_1 + a_2 = 2 + (2 + d) = 4 + d \)
- Now, calculate \( (a_1 + a_2) a_3 \):
\[
(4 + d)(2 + 2d) = (4 + d)(2 + 2d) = 8 + 8d + 2d + 2d^2 = 8 + 10d + 2d^2
\]
Finding the Minimum
To find the minimum value, we differentiate \( f(d) = 2d^2 + 10d + 8 \) and set the derivative equal to zero:
- The derivative \( f'(d) = 4d + 10 \).
Setting \( f'(d) = 0 \):
\[
4d + 10 = 0 \implies d = -\frac{10}{4} = -2.5
\]
Since \( d \) must be positive, we consider the behavior of the function for \( d > 2 \).
Conclusion
Given that \( y \) represents the common difference, if \( y > 2 \), we have:
\[
\text{As } y \text{ increases, } (a_1 + a_2)a_3 \text{ increases.}
\]
Thus, the minimum occurs at \( y = 2 \) or \( y = 3 \) satisfying the condition \( y > 2\).
In conclusion, the minimum of \( (a_1 + a_2) a_3 \) occurs at the critical point \( y = 2 \) and approaches higher values as \( y \) increases.
To solve the problem, we need to analyze the arithmetic progression (AP) where the first term \( a_1 = 2 \). The terms can be expressed as:
- \( a_1 = 2 \)
- \( a_2 = 2 + d \)
- \( a_3 = 2 + 2d \)
Here, \( d \) represents the common difference.
Expression for Minimization
We need to minimize the expression \( (a_1 + a_2) a_3 \):
- Calculate \( a_1 + a_2 \):
\( a_1 + a_2 = 2 + (2 + d) = 4 + d \)
- Now, calculate \( (a_1 + a_2) a_3 \):
\[
(4 + d)(2 + 2d) = (4 + d)(2 + 2d) = 8 + 8d + 2d + 2d^2 = 8 + 10d + 2d^2
\]
Finding the Minimum
To find the minimum value, we differentiate \( f(d) = 2d^2 + 10d + 8 \) and set the derivative equal to zero:
- The derivative \( f'(d) = 4d + 10 \).
Setting \( f'(d) = 0 \):
\[
4d + 10 = 0 \implies d = -\frac{10}{4} = -2.5
\]
Since \( d \) must be positive, we consider the behavior of the function for \( d > 2 \).
Conclusion
Given that \( y \) represents the common difference, if \( y > 2 \), we have:
\[
\text{As } y \text{ increases, } (a_1 + a_2)a_3 \text{ increases.}
\]
Thus, the minimum occurs at \( y = 2 \) or \( y = 3 \) satisfying the condition \( y > 2\).
In conclusion, the minimum of \( (a_1 + a_2) a_3 \) occurs at the critical point \( y = 2 \) and approaches higher values as \( y \) increases.
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A1, a2, a3 represents 1st, 2nd & 3rd terms of an AP respectively, 1st term is 2 & (a1 + a2)a3 is minimum, then the common difference is equal to y, y2, y3 ………; if y > 2 ퟑ퐳4y3y?
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A1, a2, a3 represents 1st, 2nd & 3rd terms of an AP respectively, 1st term is 2 & (a1 + a2)a3 is minimum, then the common difference is equal to y, y2, y3 ………; if y > 2 ퟑ퐳4y3y? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A1, a2, a3 represents 1st, 2nd & 3rd terms of an AP respectively, 1st term is 2 & (a1 + a2)a3 is minimum, then the common difference is equal to y, y2, y3 ………; if y > 2 ퟑ퐳4y3y? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A1, a2, a3 represents 1st, 2nd & 3rd terms of an AP respectively, 1st term is 2 & (a1 + a2)a3 is minimum, then the common difference is equal to y, y2, y3 ………; if y > 2 ퟑ퐳4y3y?.
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