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.

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Solution

First term $= a$
Common difference $= d$
So, $a_{r} = a + (r - 1) d$
$a_{5}^{2} + a_{11}^{2} = 3 {(a + 4 d)}^{2} + {(a + 10 d)}^{2} = 3 \Rightarrow a^{2} + 8 a d + 16 d^{2} + a^{2} + 20 a d + 10 {a d}^{2} = 3 2 a^{2} + 28 a d + 116 d^{2} = 3 a^{2} + 14 a d + 58 d^{2} = \frac{3}{2} - (i) a_{2} a_{14} = k (a + d) (a + 13 d) = k a^{2} + 13 a d + a d + 13 d^{2} = k a^{2} + 14 a d + 13 d^{2} = k$
$A s a^{2} + 14 d d = \frac{3}{2} - 58 d^{2}$ from $(i)$
$\frac{3}{2} - 58 d^{2} + 13 d^{2} = k \frac{3}{2} - 45 d^{2} = k \frac{3}{2} - k = 45 d^{2} d^{2} = \frac{3 - 2 k}{2 \times 45} d = \frac{\sqrt{3 - 2 k}}{3 \sqrt{10}}$
Sum of product of $1^{s t}$ and $15^{t h}$ term
$= a_{1} \times a_{14} = a (a + 14 d) = a^{2} + 14 a d = \frac{3}{2} - 58 d^{2} = \frac{3}{2} - \frac{58 (3 - 2 k)}{45 \times 2} = \frac{(3 \times 45) - (58) \times 3 + 2 k (58)}{45 \times 2} = \frac{39 + 116 k}{90}$

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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.

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.