Question

in a GP the 3rd term is 24 and the 6th term is 192 . find the 10 th term
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Answer 1

Answer:

Heya user....❤

Question :In a GP the 3rd term is 24 and the 6th term is 192 . Find the 10 th term.

Solution :

Let the 1st term & common ratio of the given G.P be a & r respectively.

Given, 3rd term = 24 & 6th term = 192

Therefore,

=> ar^2 = 24 and ar^5 = 192

=> ar^5/ ar^2 = 192/24

=> r^3 =8

=> r = 2

Putting the value of r in ar^2 = 24, we get a =6

Hence, 10th term = ar^9 = 6 * 2^9 = 3072

Hope it’s helpful........ ☆(❁‿❁)☆

Answer 2

Given ,

The third and sixth term of GP are 24 AnD 192

Thus ,

$\sf \star \: \: a {r}^{2} = 24 - - - (i)$

and

$\sf \star \: \: a {r}^{5} = 192 - - - (ii)$

Divide eq (ii) by eq (i), we get

$\sf \mapsto {r}^{3} = 8 \\ \\ \sf \mapsto r = \sqrt[3]{8} \\ \\ \sf \mapsto r = 2$

Put the value of r = 2 in eq (i) , we get

$\sf \mapsto a {(2)}^{2} = 24 \\ \\ \sf \mapsto4a = 24 \\ \\ \sf \mapsto a = \frac{24}{4} \\ \\ \sf \mapsto a = 6$

Now , the tenth term of GP will be

$\sf \mapsto a_{10} = 6 \times {(2)}^{9} \\ \\ \sf \mapsto a_{10} =3072$

$\therefore \sf \underline{The \: tenth \: term \: of \: GP \: is \: 3073</p><p>}$