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Find the G.P in which the \[{10^{th}}\]term is \[320\] and the \[{6^{th}}\] term is \[20\].

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Last updated date: 15th Jul 2024
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Answer
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Hint: Geometric progression is denoted by G.P. In a sequence, if each succeeding term is a multiple of each predecessor, then that sequence is known as Geometric progression. The multiple is known as the common ratio. Let \[{a_1},{a_2},{a_3},...,{a_n}\] be a sequence which is a geometric progression, then the common ratio is given by, \[\dfrac{{{a_{k + 1}}}}{{{a_k}}} = r\].

Complete step-by-step solution:
It is given that the \[{10^{th}}\] term is \[320\] and the \[{6^{th}}\] term is \[20\].
Our aim is to find a geometric progression that satisfies the given hypothesis.
We know that a geometric progression can be written as, \[{a_1},{a_2},{a_3},...,{a_n}\] where the the common ratio is given by, \[\dfrac{{{a_{k + 1}}}}{{{a_k}}} = r\].
Here, we don’t know the sequence, we only know the \[{10^{th}}\] term and the \[{6^{th}}\] term. We have to find the sequence.
We know that, the \[{n^{th}}\] term in the geometric progression is given by, \[{a_n} = a{r^{n - 1}}\],
where \[a\] is the first term of G.P and \[r\] is the common ratio.
From the question we have \[{10^{th}}\] term is \[320\].
Thus, \[{a_{10}} = a{r^{10 - 1}}\]
\[ \Rightarrow {a_{10}} = a{r^9}\]
\[ \Rightarrow a = \dfrac{{{a_{10}}}}{{{r^9}}}\]
Since, \[{a_{10}} = 320\] we get
\[a = \dfrac{{320}}{{{r^9}}}\] …………….. (A)
Likewise, it is given that \[{6^{th}}\] term is \[20\].
Thus, \[{a_6} = a{r^{6 - 1}}\].
We know that from equation (A) we get,
\[ \Rightarrow {a_6} = \dfrac{{320{r^5}}}{{{r^9}}}\]
On simplifying this we get,
\[ \Rightarrow {a_6} = \dfrac{{320}}{{{r^4}}}\]
We know that, \[{a_6} = 20\] then we get,
\[ \Rightarrow 20 = \dfrac{{320}}{{{r^4}}}\]
\[ \Rightarrow {r^4} = \dfrac{{320}}{{20}}\]
On simplifying this we get,
\[ \Rightarrow {r^4} = 16\]
Let us rewrite \[16\] as \[{2^4}\].
\[ \Rightarrow {r^4} = {2^4}\]
\[ \Rightarrow r = 2\]
Thus, we get the common ratio, \[r = 2\].
Substituting this in equation (A) we get,
\[a = \dfrac{{320}}{{{2^9}}}\]
On simplifying this we get,
\[a = \dfrac{{320}}{{512}}\]
\[\Rightarrow a = \dfrac{5}{8}\]
Thus, we got the first term of geometric progression, \[a = \dfrac{5}{8}\].
Now let’s generate the G.P using the first term \[a = \dfrac{5}{8}\] and the common ratio \[r = 2\].
\[{a_1} = \dfrac{5}{8}\], \[{a_2} = \dfrac{5}{8} \times 2 = \dfrac{5}{4}\],\[{a_3} = \dfrac{5}{4} \times 2
= \dfrac{5}{2}\], \[{a_4} = \dfrac{5}{2} \times 2 = 5\], \[{a_5} = 5 \times 2 = 10\], \[{a_5} = 10 \times 2
= 20\],…
Thus, the geometric progression is, \[\dfrac{5}{8},\dfrac{5}{4},\dfrac{5}{2},5,10,20,...\]

Note: In geometric progression no term will be equal to zero. Here, in this problem we have used the formula of \[{n^{th}}\] term of geometric progression to find the values of the first term and the common ratio as we don’t know them. After finding them we used them in the general form of G.P to generate the sequence.