Question

In a G.P,3rd term is 24 and 6th term is 192 then common ratio is​

Answer 1

Answer:

2

Step-by-step explanation:

ar^2/ar^5 = 24/192

1/r^3 = 1/8

r^3=8

r=2

Answer 2

Answer:

In a G.P, the 3rd term exists at 24 and the 6th term exists at 192 then the common ratio of r = 2.

Step-by-step explanation:

Given:

3rd term exists 24 and the 6th term exists 192

To find:

the common ratio.

Geometric Progression (GP) exists as a variety of sequences where each succeeding period exists made by multiplying each preceding term by a fixed number, which exists named a common ratio. This progression exists even comprehended as a geometric sequence of numbers that pursue a pattern.

Step 1

Consider $a_{n}=a r^{n-1}$

then $a_{n}=n^{\text {th }}$ term of Geometric Progression

n exists the number of terms

a exists the first term

r exists the common ratio

Step 2

3rd term exists 24

i.e. $a_{3}=24$

Putting $a_{n}=24, n=3$ in $a_{n}$ formula

$&24=\mathrm{a} r^{3-1}$

$&24=a r^{2}$

$&a r^{2}=24\end{aligned}$

Similarly,

Given 6th term exists 192

i.e. $a_{6}=192$

Step 3

Putting $a_{n}=192, n=6$ in $a_{n}$ formula

$&192=a r^{6-1}$

$&192=a r^{5}$

$&a r^{5}=192\end{aligned}$

Now, our equations are

$a r^{2} &=24$

$\ a r^{5} &=192\end{aligned}$

Dividing (2) by (1),

$&\frac{\mathrm{ar}^{5}}{\mathrm{ar}^{2}}=\frac{192}{24}$

$&\frac{r^{5}}{r^{2}}=8\end{aligned}$

$r^{5-2}=8$

$r^{3}=8$

Step 4

Simplifying the above equation as

$r^{3}=2 \times 2 \times 2$

$r^{3}=(2)^{3}$

r = 2.

Therefore, the common ratio of r = 2.

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