A) 4
B) 6
C) 8
D) 10
Correct Answer: B
Solution :
\[{{S}_{2n}}=3{{S}_{n}}\] \[\Rightarrow \]\[\frac{2n}{2}\{2a+(2n-1)d\}=3\ .\ \frac{n}{2}\{2a+(n-1)d\}\] \[\Rightarrow \]\[2a=(n+1)d\] Put \[2a=(n+1)d\] in\[\frac{{{S}_{3n}}}{{{S}_{n}}}\], we get its value 6.You need to login to perform this action.
You will be redirected in
3 sec