Question

insert 5 geometric means between 32/9 and 81/2

Answer 1

Answer:

Step-by-step

Answer 2

$Let \: \pink {a} \: and \: \blue { r} \: are \\</p><p>first \:term \: and \: common \: differnce \: of \\a \: G.P$

$First \:term (a) = \frac{32}{9} \:and \\Seventh \:term \: (a_{7}) = \frac{81}{2} \: (given)$

$\implies a r^{6} = \frac{81}{2}$

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

$\implies \frac{32}{9} \times r^{6} = \frac{81}{2}$

$\implies r^{6} = \frac{81}{2}\times \frac{9}{32}$

$\implies r^{6} = \frac{3^{4}}{2}\times \frac{3^{2}}{2^{5}}$

$\implies r^{6} = \frac{3^{6}}{2^{6}}$

$\implies r = \frac{3}{2}$

$Now, a = \frac{32}{9} \:and \: r = \frac{3}{2}$

$Second \:term (a_{2}) \\= ar \\= \frac{32}{9} \times \frac{3}{2} \\= \frac{16}{3}$

$Third \:term (a_{3}) \\= a_{2}r \\= \frac{16}{3} \times \frac{3}{2} \\= 8$

$Fourth\:term (a_{4}) \\= a_{3}r \\= 8 \times \frac{3}{2} \\= 12$

$Fifth\:term (a_{5}) \\= a_{4}r \\= 12 \times \frac{3}{2} \\= 18$

$Fifth\:term (a_{6}) \\= a_{5}r \\= 18 \times \frac{3}{2} \\= 27$

Therefore.,

$\red { Required \: 5 \: geometric \:means } \\\green{ \:are \: \frac{16}{3}, 8,12,18\:and \: 27 }$

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