The first term of an arithmetic progression $a_{1}, a_{2}, a_{3},$ . . .. is equal to unity. At what value of the difference of the progression is $a_{1} a_{3} + a_{2} a_{3}$ at a minimum?

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Solution

Given $a_{1} = 1$
$a_{1} a_{3} + a_{2} a_{3} = (2 + d) (1 + 2 d)$ where $d$ is the common difference required

$(2 + d) (1 + 2 d) = d ⟹ d^{2} + 2 d + 1 = 0$

The above quadratic is minimum at $⟹ d = - 1$

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The first term of an arithmetic progression a1,a2,a3, . . .. is equal to unity. At what value of the difference of the progression is a1a3+a2a3 at a minimum?

Given a1=1a1a3+a2a3=(2+d)(1+2d) where d is the common difference required (2+d)(1+2d)=d⟹d2+2d+1=0The above quadratic is minimum at ⟹d=−1

The first term of an arithmetic progression $a_{1}, a_{2}, a_{3},$ . . .. is equal to unity. At what value of the difference of the progression is $a_{1} a_{3} + a_{2} a_{3}$ at a minimum?

Given $a_{1} = 1$
$a_{1} a_{3} + a_{2} a_{3} = (2 + d) (1 + 2 d)$ where $d$ is the common difference required

$(2 + d) (1 + 2 d) = d ⟹ d^{2} + 2 d + 1 = 0$

The above quadratic is minimum at $⟹ d = - 1$