The first term of A.p is 2 and the common difference is 3 find the sum of n term of the progression. Find the value of n for which the sum of progression is 610. Find also the least value of n for which the sum exceed 1000
My answer is queen
To find the sum of the first n terms of an arithmetic progression (A.P.), we can use the formula:
Sn = (n/2) * (2a + (n-1)d)
where:
Sn = sum of n terms
n = number of terms
a = first term
d = common difference
Let's solve the questions step by step:
1. Find the value of n for which the sum of the progression is 610:
We are given:
a = 2
d = 3
Sn = 610
Using the formula, we can substitute the given values:
610 = (n/2) * (2*2 + (n-1)*3)
Rearranging the equation:
610 = (n/2) * (4 + 3n - 3)
Simplifying further:
610 = (n/2) * (7 + 3n)
Dividing both sides by (7 + 3n):
610 / (7 + 3n) = n/2
Cross-multiplying:
2 * 610 = n * (7 + 3n)
Simplifying:
1220 = 7n + 3n^2
Rearranging the equation into a quadratic form:
3n^2 + 7n - 1220 = 0
Now, we can solve this quadratic equation for n. You can use factoring, completing the square, or the quadratic formula to find the values of n.
2. Find the least value of n for which the sum exceeds 1000:
We need to find the smallest integer value of n that makes the sum Sn greater than 1000.
Using the formula, substitute:
a = 2
d = 3
Sn = 1000
Let's set up the equation:
1000 = (n/2) * (2*2 + (n-1)*3)
Solving for n, we follow the same steps as in the previous question.
Remember to check for the closest integer value of n that exceeds the desired sum.
Hope this helps! Let me know if you need further explanation.
so you have a = 2 and d = 3
sum(n) = (n/2)(4 + (n-1)3)
= (n/2)(4 + 3n - 3)
= n(3n + 1)/2
how about
n(3n+1)/2 = 610
n(3n + 1) = 1220
3n^2 + n - 1220 = 0
(n - 20)(3n + 61) = 0
n = 20 or n is negative, which is no good
so the sum of 20 terms is 610
for the last part you need:
n(3n+1)/2 > 1000
3n^2 + n - 2000 > 0
find roots of 3n^2 + n - 2000 = 0
take the next positive whole integer, eg. if the answer is 45.2, use 46