[tex]=======================[/tex]
[tex]\begin{gathered}{\underline{\huge \mathbb{P} {\large \mathrm {ROBLEM : }}}} \\\end{gathered}[/tex]
[tex]\begin{gathered}{\underline{\huge \mathbb{A} {\large \mathrm {NSWER : }}}} \\\end{gathered}[/tex]
[tex] \qquad \qquad \large \bold{4, \underline{13}, \underline{18}, 31}[/tex]
[tex]\begin{gathered}{\underline{\huge \mathbb{S} {\large \mathrm {OLUTION : }}}} \\\end{gathered} [/tex]
As given we must find the 2 arithmetic means as given from the problem. To find the arithmetic means we must find the common difference first, as of we must use the common difference formula:
[tex] \bold{Formula:}[/tex]
[tex] \qquad \boxed{ \bold{ \: \: \: d = \frac{a_n - a_k }{n - k} \: \: \: }} \\ [/tex]
So, by substituting we get:
[tex] \sf \implies{ \: d = \frac{a_n - a_k }{n - k} } \\ [/tex]
[tex] \sf \implies{ \: d = \frac{31 - 4 }{4 - 1} } \\ [/tex]
[tex] \sf \implies{ \: d = \frac{27 }{3} } \\ [/tex]
[tex] \sf \implies{ \: \bold{ d = 9 }} \\ [/tex]
Therefore, the common difference is 9.
So, we here given that:
[tex] \leadsto \bold{ \: n = 2} \\ [/tex]
[tex] \leadsto \bold{ \: a_1 = 4} \\ [/tex]
[tex] \leadsto \bold{ \: d = 9} \\ [/tex]
So, by substituting the given values at the 2nd term we get:
[tex]\sf\implies\: a_n = a_1 + (n - 1)d \\ [/tex]
[tex]\sf\implies\: a_2 = 4 + (2 - 1)9 \\ [/tex]
[tex]\sf\implies\: a_2 = 4 + ( 1)9 \\ [/tex]
[tex]\sf\implies\: a_2 = 4 + 9 \\ [/tex]
[tex]\sf\implies\: \bold{ a_2 = 13}\\ [/tex]
So, by substituting the given values at the 3rd term we get:
[tex]\sf\implies\: a_3 = 4 + (3- 1)9 \\ [/tex]
[tex]\sf\implies\: a_3 = 4 + (2)9 \\ [/tex]
[tex]\sf\implies\: a_3 = 4 + 18 \\ [/tex]
[tex]\sf\implies\: \bold{ a_3 = 22}\\ [/tex]
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Verified answer
ARITHMETIC MEAN
[tex]=======================[/tex]
[tex]\begin{gathered}{\underline{\huge \mathbb{P} {\large \mathrm {ROBLEM : }}}} \\\end{gathered}[/tex]
[tex]=======================[/tex]
[tex]\begin{gathered}{\underline{\huge \mathbb{A} {\large \mathrm {NSWER : }}}} \\\end{gathered}[/tex]
[tex] \qquad \qquad \large \bold{4, \underline{13}, \underline{18}, 31}[/tex]
[tex]=======================[/tex]
[tex]\begin{gathered}{\underline{\huge \mathbb{S} {\large \mathrm {OLUTION : }}}} \\\end{gathered} [/tex]
As given we must find the 2 arithmetic means as given from the problem. To find the arithmetic means we must find the common difference first, as of we must use the common difference formula:
[tex] \bold{Formula:}[/tex]
[tex] \qquad \boxed{ \bold{ \: \: \: d = \frac{a_n - a_k }{n - k} \: \: \: }} \\ [/tex]
So, by substituting we get:
[tex] \sf \implies{ \: d = \frac{a_n - a_k }{n - k} } \\ [/tex]
[tex] \sf \implies{ \: d = \frac{31 - 4 }{4 - 1} } \\ [/tex]
[tex] \sf \implies{ \: d = \frac{27 }{3} } \\ [/tex]
[tex] \sf \implies{ \: \bold{ d = 9 }} \\ [/tex]
Therefore, the common difference is 9.
So, we here given that:
[tex] \leadsto \bold{ \: n = 2} \\ [/tex]
[tex] \leadsto \bold{ \: a_1 = 4} \\ [/tex]
[tex] \leadsto \bold{ \: d = 9} \\ [/tex]
So, by substituting the given values at the 2nd term we get:
[tex]\sf\implies\: a_n = a_1 + (n - 1)d \\ [/tex]
[tex]\sf\implies\: a_2 = 4 + (2 - 1)9 \\ [/tex]
[tex]\sf\implies\: a_2 = 4 + ( 1)9 \\ [/tex]
[tex]\sf\implies\: a_2 = 4 + 9 \\ [/tex]
[tex]\sf\implies\: \bold{ a_2 = 13}\\ [/tex]
So, by substituting the given values at the 3rd term we get:
[tex]\sf\implies\: a_n = a_1 + (n - 1)d \\ [/tex]
[tex]\sf\implies\: a_3 = 4 + (3- 1)9 \\ [/tex]
[tex]\sf\implies\: a_3 = 4 + (2)9 \\ [/tex]
[tex]\sf\implies\: a_3 = 4 + 18 \\ [/tex]
[tex]\sf\implies\: \bold{ a_3 = 22}\\ [/tex]
[tex]=======================[/tex]