9514 1404 393
Answer:
25 terms
Step-by-step explanation:
We can use these numbers to find the first term and the common difference. Then we can use those in the formula for the sum of terms. Finally, we can solve the inequality that makes the sum of terms greater than 1560.
__
<u>First term and common difference</u>
The general term is ...
an = a1+d(n -1)
Then the given terms are ...
a2 = a1 + d(2 -1) . . . second term
8 = a1 + d . . . . . . . . . [eq1]
a4 = a1 + d(4 -1) . . . fourth term
18 = a1 + 3d . . . . . . . [eq2]
Subtracting the first equation from the second, we get ...
[eq2] -[eq1] = (18) -(8) = (a1+3d) -(a1 +d)
10 = 2d . . . . . simplify
5 = d . . . . . . . divide by 2
Using [eq1], we can find a1:
8 = a1 + 5
3 = a1 . . . . . . subtract 5
The arithmetic progression has a first term of 3 and a common difference of 5.
__
<u>Formula for the sum of terms</u>
The generic formula for the sum of an arithmetic progression is ...
Sn = n·(a1 + (a1 +d(n -1)))/2
Sn = (n/2)(2a1 + d(n -1)) . . . . simplify a little bit
With our values for a1 and d, this becomes ...
Sn = (n/2)(2·3 +5(n -1))
Sn = (n/2)(5n +1)
__
<u>Inequality for the sum</u>
We want the sum of terms to be more than 1560, so we can write the inequality ...
(n/2)(5n +1) > 1560
In standard form, this would be ...
5n² +n -3120 > 0
The quadratic formula tells us the zeros are ...
n = (-1 ±√(1² -4(5)(-3210)))/(2·5) = (-1 ± √62401)/10
The positive zero is n ≈ 24.88.
That is, for n > 24.9, the sum will be greater than 1560.
The least number of terms for a sum greater than 1560 is 25.
__
<em>Check</em>
Check the formula for 4 terms.
3 + 8 + 13 + 18 = 42 = (4/2)(5·4 +1) = 2(21) = 42 . . . note 2nd & 4th terms
Check the sums for 24 and 25 terms.
(24/2)(5·24 +1) = 12(121) = 1452 . . . . . sum < 1560 for n = 24
(25/2)(5·25 +1) = 12.5(126) = 1575 . . . sum > 1560 for n = 25
