Question

The sum of the squares of the fifth and the eleventh term of an arithmetic progression 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.

Answer 1

First term$=a$

Common difference $=d$

So,$a_r=a+(r-1)d$

$a_5^2+a_{11}^2=3{(a+4d)}^2+{(a+10d)}^2=3\Rightarrow a^2+8ad+16d^2+a^2+20ad+10{ad}^2=32a^2+28ad+116d^2=3a^2+14ad+58d^2=\frac{3}{2}-\left(i\right)a_2{a}_{14}=k(a+d)(a+13d)=k{a}^2+13ad+ad+13d^2=k{a}^2+14ad+13d^2=k$

$As{a}^2+14dd=\frac{3}{2}-58d^2$ from $\left(i\right)$

$\frac{3}{2}-58d^2+13d^2=k\frac{3}{2}-45d^2=k\frac{3}{2}-k=45d^2{d}^2=\frac{3-2k}{2\times 45}d=\frac{\sqrt{3-2k}}{3\sqrt{10}}$

Sum of product of $1^{st}$ and ${15}^{th}$ term

$=a_1\times a_{14}=a(a+14d)=a^2+14ad=\frac{3}{2}-58d^2=\frac{3}{2}-\frac{58(3-2k)}{45\times 2}=\frac{(3\times 45)-\left(58\right)\times 3+2k\left(58\right)}{45\times 2}=\frac{39+116k}{90}$