Question

The least value of n for which the sum of the series 5 + 8 + 11 +
not less than 670 is
(i) 19
(ii) 20
(ii) 21
(iv) 22​

Answer 1

Step-by-step explanation:

The formula for the sum of a series is:

Sn = n/2*[2a1 + (n-1)*d]

Sn = 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.

hope it helps

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OR

Sn = 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

Answer 2

Answer:

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