The sum of the squares of the fifth and the eleventh term of an arithmetic progression (A.P) is $3$ and the product of the second by the fourteenth term is equal to $k$. Find the product of the first by the fifteenth term 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 $= t_2 = a + d$
$5^{th}$ term of A.P $= t_5 = a + 4d$
$11^{th}$ term of A.P $= t_{11} = a + 10d$
$14^{th}$ term of A.P $= t_{14} = a + 13d$
Given: $\;$ $\left(t_5\right)^2 + \left(t_{11}\right)^2 = 3$
i.e. $\;$ $\left(a + 4d\right)^2 + \left(a + 10d\right)^2 = 3$
i.e. $\;$ $a^2 + 16d^2 + 8ad + a^2 + 100 d^2 + 20 ad = 3 $
i.e. $\;$ $2a^2 + 116 d^2 + 28 ad = 3$ $\;\;\; \cdots \; (1)$
And $\;$ $t_2 \times t_{14} = k$
i.e. $\;$ $\left(a + d\right) \left(a + 13d\right) = k$
i.e. $\;$ $a^2 + 13ad + ad + 13 d^2 = k$
i.e. $\;$ $a^2 + 13 d^2 + 14ad = k$ $\;\;\; \cdots \; (2)$
Multiplying equation $(2)$ with $2$ gives
$2a^2 + 26d^2 + 28ad = 2k$ $\;\;\; \cdots \; (3)$
Subtracting equations $(1)$ and $(3)$ gives
$90 d^2 = 3 - 2k$
i.e. $\;$ $d^2 = \dfrac{3 - 2k}{90}$ $\;\;\; \cdots \; (4)$
To find: $\;$ $t_1 \times t_{15} = a \times \left(a + 14 d\right) = a^2 + 14ad$ $\;\;\; \cdots \; (5)$
We have from equation $(2)$, $\;$ $a^2 + 14ad = k - 13 d^2$
$\begin{aligned}
\implies t_1 \times t_{15} & = k - 13 d^2 \\\\
& = k - 13 \times \left(\dfrac{3 - 2k}{90}\right) \;\;\; \left[\text{by equation (4)}\right] \\\\
& = \dfrac{90 k - 39 + 26 k}{90} \\\\
& = \dfrac{116 k - 39}{90}
\end{aligned}$