The fourth term of an arithmetic progression (A.P) is $4$. At what value of the difference of the progression is the sum of the pairwise products of the first three terms of the progression the least?
Let the first term of the A.P be $= t_1 = a$
and the common difference be $= d$
$n^{th}$ term of A.P $= t_n = a + \left(n - 1\right) d$
$\therefore \;$ $2^{nd}$ term of A.P $= t_2 = a + d$
$3^{rd}$ term of A.P $= t_3 = a + 2d$
$4^{th}$ term of A.P $= t_4 = a + 3d$
Given: $\;$ $t_4 = a + 3d = 4$
i.e. $\;$ $a = 4 - 3d$ $\;\;\; \cdots \; (1)$
And: $\;$ Sum of pairwise products of the first three terms of the progression is the least
Let $\;$ $S =$ sum of pairwise products of the first three terms of A.P
i.e. $\;$ $S = t_1 \times t_2 + t_2 \times t_3 + t_3 \times t_1$
i.e. $\;$ $S = a \times \left(a + d\right) + \left(a + d\right) \left(a + 2d\right) + \left(a + 2d\right) \times a$
i.e. $\;$ $S = a^2 + ad + a^2 + 3ad + 2d^2 + a^2 + 2ad$
i.e. $\;$ $S = 3a^2 + 6ad + 2d^2$ $\;\;\; \cdots \; (2)$
In view of equation $(1)$, equation $(2)$ becomes
$S = 3 \times \left(4 - 3d\right)^2 + 6 \times \left(4 - 3d\right) \times d + 2 d^2$
i.e. $\;$ $S = 3 \times \left(16 + 9 d^2 - 24 d\right) + 24 d - 18 d^2 + 2 d^2$
i.e. $\;$ $S = 48 + 27 d^2 - 72 d + 24 d - 16 d^2$
i.e. $\;$ $48 - 48 d + 11 d^2$ $\;\;\; \cdots \; (3)$
For $S$ to be least, $\;$ $\dfrac{d \left(S\right)}{d \left(d\right)} = 0$
$\therefore \;$ We have from equation $(3)$,
$\dfrac{d \left(S\right)}{d \left(d\right)} = 0 - 48 + 22d = 0$
i.e. $\;$ $d = \dfrac{48}{22} = \dfrac{24}{11}$
$\therefore \;$ $S$ will be least when $d = \dfrac{24}{11}$