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.
