Question

if Sn=3n+1 and Sn-1=4n-1.then find an​

Answer 1

i)

Given

Sn = 3n +1

Therefore

S1= 3×1+1=3+1=4

S2= 3×2+1 =6+1=7

Therefore

a=4

a2= S2-a = 7-4=3

So

Common Difference d= 3-4 =-1

Therefore

an = a+(n-1)d

=4 +(n-1)×(-1)

=4 -n +1

=5 -n

2)

Sn -1 = 4n-1

➡ Sn = 4n -1+1

➡ Sn= 4n

Therefore

S1= 4×1=4

S2= 4×2= 8

Hence

a = 4

a2= S2-s1 =8-4=4

Common difference d= 4-4=0

Therefore

an= a+(n-1)d

= 4+(n-1)×4

=4 +4n-4

=4n