Question

Help me to solve this question. ​

Answer 1

Answer:

  n = 133

Step-by-step explanation:

The sum of n terms of an arithmetic sequence is ...

  Sn = n(2·a1 +d(n -1))/2

You want ...

  n(10 +1/2(n -1))/2 > 5000 . . . . . filling in a1=5, d=1/2

  (n^2 +19n)/4 > 5000 . . . . . . . . .simplify

  n^2 +19n -20000 > 0  . . . . . . . write in standard form

We can solve this quadratic using the quadratic formula ...

  n > (-19 +√(19^2 -4(1)(-20000)))/2 = (-19 +√80361)/2 ≈ 132.24

The least value of n such that the sum of n terms exceeds 5000 is ...

  n = 133.