1) If the sum of the first n terms of a series is Sn where Sn=2n²-n,
(a)prove that the series is an AP, stating the first term and the common difference;
(b)find the sum of the terms from the 3rd to the 12th inclusive
2) In an AP the 6th term is half the 4th term and the 3rd term is 15.
(a)Find the first term and the common difference;
(b)How many terms are needed to give a sum that is less than 65?
Please help
Sn=2n+2-4
Sure, I can help you with these questions. Let's tackle them one by one.
Question 1:
(a) To prove that the series is an arithmetic progression (AP), we need to show that the difference between consecutive terms is constant.
Let's start by finding the general expression for the nth term of the series.
The sum of the first n terms, Sn, is given by Sn = 2n² - n.
To find the nth term, we can use the formula for the sum of an arithmetic series:
Sn = (n/2)(2a + (n-1)d), where a is the first term and d is the common difference.
Comparing this with the given expression for Sn = 2n² - n, we can equate the corresponding terms:
2n² - n = (n/2)(2a + (n-1)d)
Expanding the right side of the equation, we get:
2n² - n = an + (n² - n)(d/2)
2n² - n = an + (nd² - nd - d/2)
2n² - n = an + nd² - nd - d/2
Now, let's equate the coefficients of the like terms:
-1 = -d/2
Simplifying this equation, we find: d = 2.
Now that we know the common difference, we can substitute it back into the equation to find the first term, a:
2n² - n = an + nd² - nd - d/2
2n² - n = an + 2n² - 2n - 1
1 = an - 2n - 1
From this equation, we can deduce that an - 2n = 2.
Substituting n = 1, we find: a - 2 = 2
Simplifying this, we get: a = 4.
Therefore, the series is an AP with the first term, a, equal to 4 and the common difference, d, equal to 2.
(b) To find the sum of the terms from the 3rd to the 12th inclusive, we can use the formula for the sum of an arithmetic series:
Sn = (n/2)(2a + (n-1)d)
Plugging in the values, we get:
S12 = (12/2)(2*4 + (12-1)*2)
= 6(8 + 22)
= 6(30)
= 180.
Therefore, the sum of the terms from the 3rd to the 12th inclusive is 180.
Question 2:
(a) Let's assume that the first term of the arithmetic progression is a and the common difference is d.
Given that the 6th term is half the 4th term, we can write:
a + 5d = (a + 3d)/2
Now let's solve this equation to find the value of d:
2(a + 5d) = a + 3d
2a + 10d = a + 3d
a = -7d
Given that the 3rd term is 15, we can substitute this into the equation:
a + 2d = 15
-7d + 2d = 15
-5d = 15
d = -3
Now we can find the value of a:
a = -7d = -7(-3) = 21
Therefore, the first term is 21 and the common difference is -3.
(b) To find the number of terms needed to give a sum less than 65, we can use the formula for the sum of an arithmetic series:
Sn = (n/2)(2a + (n-1)d)
We can rewrite this as:
Sn = (n/2)(a + l), where l is the last term of the series.
We want to find the number of terms, n, such that Sn < 65.
Plugging in the values, we get:
(n/2)(a + a + (n-1)d) < 65
(n/2)(2a + (n-1)d) < 65
Substituting the values of a and d from part (a), we have:
(n/2)(2(21) + (n-1)(-3)) < 65
(n/2)(42 - 3n + 3) < 65
(n/2)(45 - 3n) < 65
45n - 3n² < 130
Simplifying this equation, we have:
-3n² + 45n - 130 < 0
To solve this quadratic inequality, we can find the roots of the corresponding quadratic equation:
-3n² + 45n - 130 = 0
Using the quadratic formula, we find:
n = (-45 ± √(45² - 4(-3)(-130))) / (2(-3))
n = (-45 ± √(2025 - 1560)) / (-6)
n = (-45 ± √465) / (-6)
Calculating further:
n ≈ (-45 ± 21.57) / (-6)
n ≈ (45 - 21.57) / 6 (choosing the positive root)
n ≈ 23.43 / 6
n ≈ 3.90
Since the number of terms must be a positive integer, we can round up to n = 4.
Therefore, we need at least 4 terms in order to have a sum less than 65.