Question

in a GP 3rd term is 24 and 6th term is 192 then 10th term is ​

Answer 1

Answer:

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

Dividing,

r^3 = 192/24 = 8

r = 2

Therefore, a = 6

10th term = ar^9 = 6*2^9 = 6*2^10 = 6*1024 = 6144

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Answer 2

Concept Introduction:

'GP' is the short term of geometric progression. In mathematics, G.P. is a sequence where each term is produced by multiplying a preceding term by a fixed number.

Given:

We have been given a question in which a G.P. series has 24 in its 3rd term and 192 in the 6th term.

To Find:

We have to find the 10th term in the G.P. series.

Solution:

According to the problem

$\begin{array}{l}a_{n}=a r^{n-1} \\a_{3}=a r^{2}=24 \ldots(1) \\a_{6}=a r^{5}=192 \ldots(2)\end{array}\\dividing (2) and (1), we get\\\begin{array}{l}\frac{a r^{5}}{a r^{2}}=\frac{192}{24} \\r^{3}=8 \ , r=2\end{array}\\from\left(1)\\ar^{2}=24 a=\frac{24}{2^{2}}=6 \therefore a_{10}=(6)(2)^{10-1}=3072$

Final Answer:

3072 is the 10th term.

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