Question

If Sn=n(4n+1) find Ap

Answer 1

➡An=Sn-Sn-1

➡An=n(4n+1)-[(n-1){4(n-1)+1}]

➡An=4n^2+n-[(n-1){4n-4+1}]

➡An=4n^2+n-[(n-1){4n-3}]

➡An=4n^2+n-[4n^2-3n-4n+3]

➡An=4n^2+n-4n^2+7n-3

➡An=8n-3

➡A1=8-3=5

➡A2=8×2-3=13

➡A3=8×3-3=21

Ap➡5,13,21...........with common difference 8

Answer 2

sn=n (4n+1)
 
put n=1

s1=1(4x1+1)
    =4+1
    =5

a=5
 
put n=2

s2= 2(4x2+1)
    =2x9 
    =18

a1+a2=18
5+a2=18
a2=13

d=a1-a2
 =13-5
 =8
    
a3=a+(n-1)d
    =5+3-1x8
    =5+16
    =21
       
so AP is 5,13,21.....