Answer: (a) a = 3, and d=[tex]\dfrac{5}{2}[/tex]
(b) [tex]a_4=\dfrac{21}{2}\\\\a_5=13\\\\a_6=\dfrac{31}{2}[/tex]
Step-by-step explanation:
Since we have given that
Arithmetic sequence be
[tex]3,\dfrac{11}{2},8[/tex]
As we know that
a denotes the first term.
d denotes the common difference.
So, [tex]a^1=3\\\\d=a_2-a_1=\dfrac{11}{2}-3=\dfrac{11-6}{2}=\dfrac{5}{2}[/tex]
Hence, a = 3, and d=[tex]\dfrac{5}{2}[/tex]
[tex]a_4=a+3d=3+3\times\dfrac{5}{2}=3+\dfrac{15}{2}=\dfrac{6+15}{2}=\dfrac{21}{2}\\\\a_5=a+4d=3+4\times \dfrac{5}{2}=3+\dfrac{20}{2}=3+10=13\\\\a_6=a+5d=3+5\times \dfrac{5}{2}=3+\dfrac{25}{2}=\dfrac{6+25}{2}=\dfrac{31}{2}[/tex]
Hence, (b)
[tex]a_4=\dfrac{21}{2}\\\\a_5=13\\\\a_6=\dfrac{31}{2}[/tex]
