Here , we need to find a formula for nth term of the sequence 6 , 12 , 18 , .....
Now , here if you notice carefully ,then you can notice that 12 - 6 = 6 and 18 - 12 = 6 , So the given sequence is an AP ( Arithmetic Progression ) with common difference being 6 , first term being 6 , so now as we know that :
- [tex]{\boxed{\bf{a_{n}=a+(n-1)d}}}[/tex]
Where , [tex]\bf a_n[/tex] is nth term of AP , [tex]\bf a[/tex] being first term while [tex]\bf d[/tex] is the common difference . So , now putting the values in above formula ;
[tex]{:\implies \quad \sf a_{n}=6+(n-1)6}[/tex]
[tex]{:\implies \quad \sf a_{n}=6+6n-6}[/tex]
Cancelling 6 will yield :
[tex]{:\implies \quad \bf \therefore \quad \underline{\underline{a_{n}=6n}}}[/tex]
Henceforth , nth term of the sequence is 6n
