In a g.p the 5th term is 48 and 2nd term is 6 . find the common ratio

Answer:

[tex]\sf{ \:\qquad\qquad\qquad \boxed{ \sf{ \:\bf \: r = 2 \: }}} \\ \\ [/tex]

Step-by-step explanation:

Let assume that first term of a GP series is a and common ratio of GP series is r.

Given that,

[tex]\qquad\sf \:a_5 = 48 \\ \\ [/tex]

[tex]\qquad\sf \: {ar}^{5 - 1} = 48 \\ \\ [/tex]

[tex]\qquad\sf \: {ar}^{4} = 48 \: \: - - - (1) \\ \\ [/tex]

Further, given that

[tex]\qquad\sf \:a_2 = 6 \\ \\ [/tex]

[tex]\qquad\sf \: {ar}^{2 - 1} = 6 \\ \\ [/tex]

[tex]\qquad\sf \: ar = 6 \: - - - (2)\\ \\ [/tex]

On dividing equation (2) by (1), we get

[tex]\qquad\sf \:\dfrac{ {ar}^{4} }{ar} = \dfrac{48}{6} \\ \\ [/tex]

[tex]\qquad\sf \: {r}^{3} = 8 \\ \\ [/tex]

[tex]\qquad\sf \: {r}^{3} = {2}^{3} \\ \\ [/tex]

[tex]\qquad\sf\implies \sf \: r = 2 \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

↝ Sum of n  terms of an geometric progression is,

[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{S_n\:=\dfrac{a( {r}^{n} - 1) }{r - 1} }}}}}} \\ \end{gathered} \\ \\ [/tex]

↝ nᵗʰ term of an geometric progression is,

[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{a_n\:=\: {ar}^{n - 1} }}}}}} \\ \end{gathered} \\ \\ [/tex]

Wʜᴇʀᴇ,

a is the first term of the progression.

n is the no. of terms.

r is the common ratio.

Answer:

The common ration = 2

Step-by-step explanation:

In a geometric progression, the nth term is given by:

[tex]t_{n} = ar^{n-1}[/tex]

The common ration is given by:

[tex]r = \frac{t^{n}}{t^{n-1}}[/tex]

We have been given,

[tex]t_{5} = 48[/tex]

[tex]t_{2} = 6[/tex]

Which can be written as,

[tex]ar^{4} = 48[/tex]

[tex]ar^{1} = 6[/tex]

Step 2:

Taking the ration of the 5th term to 2nd term, we get,

[tex]\frac{t_{5}}{t_{2}} = \frac{ar^{4} }{ar^{1} } = \frac{48}{6}[/tex]

r³ = 8

∴ r = 2

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