5+8+11+14+...=670

Question

Find how many terms of this arithmetic series are needed to make the given sum

Answer 1

The formula for the sum of a series is:

 

S_n = n/2*[2a₁ + (n-1)*d]

 

S_n = 670, a₁ = 5, d = 3, so:

 

670 = n/2*[2*5 + (n-1)*3]

670 = n/2*[10 + 3n - 3]

1340 = 3n² + 7n

3n² + 7n -1340 = 0

 

Plug that in to the quadratic equation and we get:

 

n = [-7 ± sqrt(49 - 4*3*-1340)]/2*3

n = [-7 ± sqrt(49 + 16080)]/6

n = [-7 ± sqrt(16129)]/6

n = [-7 ± 127]/6

n = 120/6 or -134/6

since n terms can only be positive, n = 20.

 

Answer 2

Hi Sharik,

 

S_n = n/2[2a + (n-1)d]

 

Sn = 670, a=5, d= 3

 

670 = n/2 [10+(n-1)3]

 

670* 2 = n[10+3n-3]

 

1340 = 7n +3n²  

 

0 = 3n² +7n -1340

 

0 = 3n² +67n-60n-1340

 

0= n(3n+67) -20(3n+67)

 

0=(3n+67)(n-20)

 

so 3n+67=0 or n-20=0

 

n=-67/3 0r n=20

 

Ans: No of terms = 20