Question

If the 5th term, 8th term of G.P 2power0,2power1,2power2 ......are roots of quadratic equation then find quadratic equation​

Answer 1

$Given \:G.P : 2^{0}, 2^{1} , 2^{3},\cdot \cdot\cdot ,$

$First \:term (a) = 2^{0} = 1$

$Common \:ratio (r) = \frac{a_{3}}{a_{2}}\\= \frac{2^{3}}{2^{2}} \\= \frac{8}{4} \\= 2$

$we \:know \:that ,$

$\boxed {\pink { n^{th} \:term (a_{n}) = ar^{n-1}}}$

$i) 5^{th} \:term = a_{5} \\= ar^{5-1}\\= ar^{4} \\= 1 \times 2^{4} \\= 16 \:---(1)$

$ii) 8^{th} \:term = a_{8} \\= ar^{8-1}\\= ar^{7} \\= 1 \times 2^{7} \\= 128 \:---(2)$

$Now, If \:a_{5} = 16\:and \:a_{8} = 128\:are\\ \:roots\:of \: Quadratic \:equation$

$Sum \:of \:the \:roots = 16 + 128 = 144$

$Product \:of \:the \:roots = 16 \times 128 \\= 2048$

$Required \: Quadratic \: Equation : \\</p><p>x^{2} - (sum \:of \:roots )x + product \:of \:roots = 0$

$\implies\green { x^{2} - 144x + 2048 = 0}$

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