SOLUTION: An arithmetic progression has a second term of 8 and a fourth term of 18. Find the least number of terms in which the sum of their progression is greater than 1560

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Question 1206121: An arithmetic progression has a second term of 8 and a fourth term of 18. Find the least number of terms in which the sum of their progression is greater than 1560
Answer by math_tutor2020(3816) About Me  (Show Source):
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d = common difference
2nd term = 8
3rd term = (2nd term) + d = 8 + d
4th term = (3rd term) + d = (8+d) + d = 8 + 2d
4th term = 18

8+2d = 18
2d = 18-8
2d = 10
d = 10/2
d = 5

2nd term = 8
3rd term = 8 + d = 8 + 5 = 13
4th term = 8 + 2d = 8+2*5 = 18

In short,
1st term = 3
2nd term = 8
3rd term = 13
4th term = 18
etc.
The pattern is that the units digit alternates 3,8,3,8,...

a = first term = 3
d = common difference = 5
n = number of terms
Sn = sum of the first n terms of an arithmetic sequence
Sn = (n/2)*(2a + d*(n-1))
Sn = (n/2)*(2*3 + 5*(n-1))
Sn = 2.5n^2 + 0.5n

Sn > 1560
2.5n^2 + 0.5n > 1560
2.5n^2 + 0.5n - 1560 > 0

Let's solve an adjacent or similar equation
2.5x^2 + 0.5x - 1560 = 0
Multiply both sides by 2 to end up with
5x^2 + x - 3120 = 0

Then let's use the quadratic formula.
a = 5, b = 1, c = -3120
x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

x+=+%28-1%2B-sqrt%28%281%29%5E2-4%285%29%28-3120%29%29%29%2F%282%285%29%29

x+=+%28-1%2B-sqrt%2862401%29%29%2F%2810%29

x+=+%28-1%2Bsqrt%2862401%29%29%2F%2810%29 or x+=+%28-1-sqrt%2862401%29%29%2F%2810%29

x+=+%28-1%2B249.801922%29%2F%2810%29 or x+=+%28-1-249.801922%29%2F%2810%29

x+=+%28248.801922%29%2F%2810%29 or x+=+%28-250.801922%29%2F%2810%29

x+=+24.880192 or x+=+-25.080192
The decimal values are approximate.
Ignore the negative solution since n > 0

Since x = 24.880192 is one approximate solution to 2.5x^2 + 0.5x - 1560 = 0, let's try n = 24 to see what happens to the arithmetic sum.
Sn = 2.5n^2 + 0.5n
S24 = 2.5*24^2 + 0.5*24
S24 = 1452
The sum of the first 24 terms is 1452.
This is too small since we need something larger than 1560.

Now try n = 25
Sn = 2.5n^2 + 0.5n
S25 = 2.5*25^2 + 0.5*25
S25 = 1575
The sum of the first 25 terms is 1575.
We have cleared over the hurdle.

Therefore, you need at least 25 terms added up to get an arithmetic sum larger than 1560.