Answer:
Proved below
Step-by-step explanation:
Given that
[tex]5a_{5}=8a_{8}[/tex] .......(1
as we know
[tex]a_{5}= a_{1} + 4d[/tex]
[tex]a_{8}= a_{1} + 7d[/tex]
where d is the common difference. So, equation (1) would become
[tex]5(a_{1} + 4d)=8(a_{1} + 7d)[/tex]
[tex]5a_{1} + 20d=8a_{1} + 56d[/tex]
Rearranging
[tex]5a_{1} - 8a_{1}=-20d + 56d[/tex]
[tex]- 3a_{1}=36d[/tex]
So,
[tex]a_{1}=-12d[/tex] .....(2)
As we know
[tex]a_{13}= a_{1} + 12d[/tex] ......(3)
So, substituting the value of a1 into equation (3)
[tex]a_{13}= -12d + 12d[/tex]
[tex]a_{13}= 0[/tex]
Hence Proved
