Question

Consider the arithmetic series 4+7+10+13+16+19.... Find n such that Sn = 175

Answer 1

Step-by-step explanation:

Sn = 175

Sn = 175a=4

Sn = 175a=4d=7-4=3

Sn = 175a=4d=7-4=3n/2(2a+(n-1)d) =Sn

Sn = 175a=4d=7-4=3n/2(2a+(n-1)d) =Snn/2(2(4)+(n-1)3)=175

Sn = 175a=4d=7-4=3n/2(2a+(n-1)d) =Snn/2(2(4)+(n-1)3)=175n/2(8+3n-3)=175

Sn = 175a=4d=7-4=3n/2(2a+(n-1)d) =Snn/2(2(4)+(n-1)3)=175n/2(8+3n-3)=175n/2(3n+5)=175

Sn = 175a=4d=7-4=3n/2(2a+(n-1)d) =Snn/2(2(4)+(n-1)3)=175n/2(8+3n-3)=175n/2(3n+5)=1753n^2+5n/2=175

Sn = 175a=4d=7-4=3n/2(2a+(n-1)d) =Snn/2(2(4)+(n-1)3)=175n/2(8+3n-3)=175n/2(3n+5)=1753n^2+5n/2=1753n^2+5n=175×2

Sn = 175a=4d=7-4=3n/2(2a+(n-1)d) =Snn/2(2(4)+(n-1)3)=175n/2(8+3n-3)=175n/2(3n+5)=1753n^2+5n/2=1753n^2+5n=175×23n^2+5n=350

Sn = 175a=4d=7-4=3n/2(2a+(n-1)d) =Snn/2(2(4)+(n-1)3)=175n/2(8+3n-3)=175n/2(3n+5)=1753n^2+5n/2=1753n^2+5n=175×23n^2+5n=3503n^2+5n-350=0

Sn = 175a=4d=7-4=3n/2(2a+(n-1)d) =Snn/2(2(4)+(n-1)3)=175n/2(8+3n-3)=175n/2(3n+5)=1753n^2+5n/2=1753n^2+5n=175×23n^2+5n=3503n^2+5n-350=03n^2-30n+35n-350=0

Sn = 175a=4d=7-4=3n/2(2a+(n-1)d) =Snn/2(2(4)+(n-1)3)=175n/2(8+3n-3)=175n/2(3n+5)=1753n^2+5n/2=1753n^2+5n=175×23n^2+5n=3503n^2+5n-350=03n^2-30n+35n-350=03n(n-10)+35(n-10)=0

(3n+35)(n-10)=0

3n+35=0

3n=-35

n-10=0

n=10

n can't be negative