How Mandy terms in the séries 19+16+13.......must ne aide to give sur of 44
The given series is an arithmetic progression with a common difference of -3 (16-19 = -3, 13-16 = -3, ...).
To find the number of terms needed to sum to 44, we can use the formula for the sum of an arithmetic series:
Sn = n/2 * (2a + (n-1)d)
Where:
Sn = sum of the first n terms
n = number of terms
a = first term
d = common difference
We are given that Sn = 44 and a = 19, d = -3. Substituting these values into the formula:
44 = n/2 * (2*19 + (n-1)*(-3))
44 = n/2 * (38 - 3n + 3)
44 = n/2 * (41 - 3n)
Multiplying both sides by 2 to get rid of the division:
88 = n * (41 - 3n)
88 = 41n - 3n^2
Rearranging the equation to set it equal to zero:
3n^2 - 41n + 88 = 0
Now, we can find the values of n that satisfy this quadratic equation:
Using the quadratic formula:
n = (-(-41) ± √((-41)^2 - 4*3*88)) / 2*3
n = (41 ± √(1681 - 1056)) / 6
n = (41 ± √625) / 6
n = (41 ± 25) / 6
n = (41 + 25) / 6 = 66 / 6 = 11
n = (41 - 25) / 6 = 16 / 6 = 8.66 (approximately)
Since the number of terms must be a whole number, the series must have 11 terms to sum to 44.