Question

find GP whose 5 term is 80 and 8 term is 640

Answer 1

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$\large\mathcal\red{solution}$

Now ,let the 1st term is= a and common ratio is =r

now condition 1

$a(5) = ar {}^{5 - 1} = ar {}^{4} = 80 \\ and</u></strong><strong><u> </u></strong><strong><u>\</u></strong><strong><u>:</u></strong></p><p><strong><u>condit</u></strong><strong><u>ion</u></strong><strong><u> </u></strong><strong><u>2</u></strong><strong><u> \\ a(8) = ar {}^{8 - 1} = ar {}^{7} = 640 \\ now \\ \frac{a(5)}{a(8)} = \frac{80}{640} = \frac{ar {}^{4} }{ar {}^{7} } = \frac{1}{r {}^{3} } \\ = > r {}^{3} = \frac{640}{80} \\ = > r = 2 \\ now \: \\ a = \frac{80}{2 {}^{4} } = 5 \\ \\ therefore \: the \: series \: is \:$

5,10,20,40,80,160,320,640,1280,...............