The sum of the first and the fifth term of an arithmetic progression (A.P) is $26$ and the product of the second by the fourth term is $160$. Find the sum of the first six terms of the progression.
Let the first term of A.P $= t_1 = a$
Let the common difference of A.P $= d$
$n^{th}$ term of A.P $= t_n = a + \left(n - 1\right) d$
$\therefore \;$ $2^{nd}$ term of A.P $= t_2 = a + d$
$4^{th}$ term of A.P $= t_4 = a + 3d$
$5^{th}$ term of A.P $= t_5 = a + 4d$
Given: $\;$ $t_1 + t_5 = 26$
i.e. $\;$ $a + a + 4d = 26$
i.e. $\;$ $2a + 4d = 26$
i.e. $\;$ $a + 2d = 13$
i.e. $\;$ $a = 13 - 2d$ $\;\;\; \cdots \; (1)$
And $\;$ $t_2 \times t_4 = 160$
i.e. $\;$ $\left(a + d\right) \left(a + 3d\right) = 160$ $\;\;\; \cdots \; (2)$
In view of equation $(1)$ equation $(2)$ becomes
$\left(13 - 2d + d\right) \left(13 - 2d + 3d\right) = 160$
i.e. $\;$ $\left(13 - d\right) \left(13 + d\right) = 160$
i.e. $\;$ $169 - d^2 = 160$
i.e. $\;$ $d^2 = 9$ $\implies$ $d = \pm 3$
When $\;$ $d = +3$, $\;$ we have from equation $(1)$, $\;$ $a = 13 - 6 = 7$
When $\;$ $d = -3$, $\;$ we have from equation $(1)$, $\;$ $a = 13 + 6 = 19$
Sum of $n$ terms of A.P $= \dfrac{n \left[2a + \left(n - 1\right) d\right]}{2}$
$\therefore \;$ Sum of first six terms of A.P $= S_6 = \dfrac{6 \times \left[2a + 5d\right]}{2} = 3 \left[2a + 5d\right]$
Case 1: $\;$ When $a = 7, \; d = +3$, $\;$ $S_6 = 3 \times \left[2 \times 7 + 5 \times 3\right] = 87$
Case 2: $\;$ When $a = 19, \; d = -3$, $\;$ $S_6 = 3 \times \left[2 \times 19 - 5 \times 3\right] = 69$