Plz the first and last term of an AP are 79 and -5 if the sum is 814. Find
A. The number of terms
B. Common difference.
http://www.mathsisfun.com/algebra/sequences-sums-arithmetic.html
XN = 79+d(N-1) = -5
814 = N/2[ 2(79) + (N-1)d ]
814 = N/2 [ 79 -5 ]
N = 1628/74
N= 22
then
-5 = 79 +d(21)
d = -84/21 = - 4
where is the last answer
A. Well, the number of terms in this arithmetic progression can be determined by finding the difference between the first and last term and then dividing it by the common difference. So, let's put on our arithmetic goggles and do some math!
79 - (-5) = 84
Since the sum of an AP is given by the formula (n/2) * (2a + (n-1)d), we can plug in the given values:
814 = (n/2) * (2 * 79 + (n-1) * d)
That's a lot of variables! But fear not, we can simplify this equation by substituting the values that we already know:
814 = (n/2) * (158 + (n-1) * d)
Now let's solve for n. Since we're a clown bot, I'll do a funny dance while you follow along:
814 * 2 = n * (158 + (n-1) * d)
1628 = n * (158 + nd - d)
1628 = 158n + n²d - nd
And now let's put it in standard quadratic form by rearranging:
n²d - (d - 158)n + 1628 = 0
But wait! We don't know the common difference yet, so we can't solve this equation just yet. Keep reading for Part B to find out!
B. We'll need to use the fact that the sum is given by the formula (n/2) * (2a + (n-1)d), but luckily we already know it equals 814. So let's substitute that in!
814 = (n/2) * (2 * 79 + (n-1) * d)
Now let's simplify this equation:
814 = (n/2) * (158 + (n-1) * d)
814 = (n * 158)/2 + (n² * d - nd)/2
814 = 79n + (n²d - nd)/2
But uh-oh, we have another equation with too many variables. We need to know either the number of terms (n) or the common difference (d) to find the other. Unfortunately, we're stuck in a loop of uncertainty, so we can't find either the number of terms or the common difference without the other value. It seems like this arithmetic progression has gotten itself into some mathematical clownery!
To find the number of terms and the common difference of an arithmetic progression (AP), we'll use the following formulas:
1. The nth term of an AP: \(a_n = a_1 + (n-1)d\)
2. The sum of the first n terms of an AP: \(S_n = \frac{n}{2}(2a_1 + (n-1)d)\)
Given information:
First term, \(a_1 = 79\)
Last term, \(a_n = -5\)
Sum of the AP, \(S_n = 814\)
Now, let's find the number of terms (n) and the common difference (d):
A. Number of terms (n):
We know that \(a_n = a_1 + (n-1)d\), so we can substitute the values:
\(-5 = 79 + (n-1)d\)
Simplify the equation:
\(-5 - 79 = (n-1)d\)
\(-84 = (n-1)d\)
B. Common difference (d):
Next, we'll use the formula for the sum of the first n terms. We'll substitute the known values:
\(S_n = \frac{n}{2}(2a_1 + (n-1)d)\)
\(814 = \frac{n}{2}(2(79) + (n-1)d)\)
\(814 = \frac{n}{2}(158 + (n-1)d)\)
We now have two equations:
Equation 1: \(-84 = (n-1)d\)
Equation 2: \(814 = \frac{n}{2}(158 + (n-1)d\)
To solve these equations simultaneously, we can use substitution or elimination. Let's use substitution.
First, solve Equation 1 for \(d\):
\(d = \frac{-84}{n-1}\)
Now, substitute this value of \(d\) into Equation 2:
\(814 = \frac{n}{2}(158 + (n-1)\left(\frac{-84}{n-1}\right)\)
Simplify the equation:
\[814 = \frac{n}{2}(158 - 84)\]
\[814 = \frac{n}{2}(74)\]
Multiply both sides by 2 to remove the fraction:
\[1628 = n \cdot 74\]
Now solve for \(n\):
\[n = \frac{1628}{74} = 22\]
To find the common difference (d), substitute the value of \(n\) in Equation 1:
\[d = \frac{-84}{n-1} = \frac{-84}{22-1} = -4\]
Therefore,
A. The number of terms is 22.
B. The common difference is -4.