Question

In a certain A.P. the $24th$ term is twice the $10th$ term. Prove that the $36th$ term is twice the $16th$ term.

Answer 1

$\because a_{24}=2.a_{10}\Rightarrow a+(24-1)d=2(a+(10-1\left)d\right)\Rightarrow a+23d=2a+18d\Rightarrow a=5d⟶\left(1\right)Now,a_{36}=a+(36-1)d\Rightarrow a_{36}=a+35d\Rightarrow a_{36}=a+5d+30d\Rightarrow a_{36}=a+a+30d\left(from\right(1\left)\right)\Rightarrow a_{36}=2a+30d\Rightarrow a_{36}=2(a+15d)\Rightarrow a_{36}=2a_{16}$