We are to find the number of terms in the AP that has a sum greater than 250
The number of teams in the AP which has a sum greater than 250 is at least 11 terms
First term, a = 3
second term, a + d = 8
so
common difference, d = 8-3
= 5
let
n = number of terms in the AP
The sum of n terms is 250
Sum = n/2{2a + (n - 1)d}
250 = n/2{2*3 + (n-1)5 }
250 = n/2{6 + 5n - 5}
250 = n/2(1 + 5n)
500 = n + 5n²
5n² + n - 500 = 0
Solve quadratically using formula
n = -b ± √b² - 4ac / 2a
= -1 ± √1² - 4(5)(-500) / 2(5)
= -1 ± √1 - (-10000) / 10
= -1 ± √1 + 10000 / 10
= -1 ± √10001 / 10
= -1 + √10001 / 10 or -1 - √10001 / 10
= -1/10 + √10001/10
or
-1/10 - √10001 / 20
= 9.9005 or -10.1005
The value of n must be positive and a whole value
Therefore,
n = 9.9005
Approximately, n = 10
The number of teams in the AP which has a sum greater than 250 is at least 11 terms
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