If the third term of a linear sequence is 20 and the sum of the first ten terms 275. Find the first term and the common difference
Well, let's put on our math clown noses and solve this puzzle!
First, let's figure out the common difference. In a linear sequence, each term is obtained by adding the common difference to the previous one. If the third term is 20, we know that the second term is 20 minus the common difference, and the first term is 20 minus twice the common difference.
We can say that the second term is 20 - x, and the first term is 20 - 2x. The sum of the first ten terms is the sum of the terms from 1 to 10, and can be calculated using the formula: S = (n/2)(2a + (n-1)d), where S is the sum, a is the first term, d is the common difference, and n is the number of terms.
So, we have the equation:
275 = (10/2)(2(20 - 2x) + (10-1)x)
Simplifying that equation, we get:
275 = 5(40 - 4x + 9x)
Let's continue simplifying:
275 = 200 - 5x
Now, let's isolate x:
5x = 75
x = 15
Now that we know the common difference is 15, we can find the first term:
First term = 20 - 2x = 20 - 2(15) = -10
So, the first term of the linear sequence is -10 and the common difference is 15. Keep in mind, I'm just a clown bot, so please double-check my calculations!
Let's use the formula for the nth term of a linear sequence:
an = a1 + (n - 1)d,
where an represents the nth term, a1 represents the first term, n represents the term number, and d represents the common difference.
We are given that the third term (a3) is 20, so we can substitute these values into the formula:
20 = a1 + (3 - 1)d,
Simplify the equation:
20 = a1 + 2d. ...(Equation 1)
We are also given that the sum of the first ten terms (Sn) is 275. The formula for the sum of the first n terms of an arithmetic series is:
Sn = (n/2) * (a1 + an).
We can use this formula to find another equation:
275 = (10/2) * (a1 + a10).
Simplify the equation:
275 = 5 * (a1 + (a1 + 9d)),
275 = 5 * (2a1 + 9d), ...(Equation 2)
Now we have a system of two equations (Equation 1 and Equation 2) with two unknowns (a1 and d). We can solve this system of equations to find the values of a1 and d.
First, let's multiply Equation 1 by 5 to eliminate the a1 term:
100 = 5a1 + 10d. ...(Equation 3)
Now we can solve equations 2 and 3 simultaneously. Subtracting Equation 3 from Equation 2, we get:
175 = (5 * (2a1 + 9d)) - (5a1 + 10d),
175 = 10a1 + 45d - 5a1 - 10d,
175 = 5a1 + 35d.
Let's call this new equation Equation 4.
Now, subtract Equation 3 from Equation 4 to eliminate the a1 term:
175 - 100 = (5a1 + 35d) - (5a1 + 10d),
75 = 25d.
Divide both sides of the equation by 25:
75/25 = 25d/25,
3 = d.
So the common difference (d) is 3.
Now we can substitute this value of d into Equation 1 to find the first term (a1):
20 = a1 + 2(3),
20 = a1 + 6.
Subtract 6 from both sides of the equation:
20 - 6 = a1 + 6 - 6,
14 = a1.
So the first term (a1) is 14.
Therefore, the first term is 14 and the common difference is 3.
To find the first term and the common difference of a linear sequence given the third term and the sum of the first ten terms, we can use these steps:
Step 1: Find the common difference (d)
To find the common difference (d), we can use the formula for the nth term of a linear sequence:
n-th term = a + (n - 1) * d
where "a" is the first term, "n" is the position of the term, and "d" is the common difference.
We are given that the third term (n = 3) is 20:
20 = a + 2d ...(1)
Step 2: Find the sum of the first ten terms (S10)
The sum of the first ten terms (S10) of a linear sequence can be calculated using the formula:
S10 = (n / 2) * (2a + (n - 1) * d)
where "n" is the number of terms, "a" is the first term, and "d" is the common difference.
We are given that the sum of the first ten terms (S10) is 275:
275 = (10 / 2) * (2a + (10 - 1) * d)
= 5 * (2a + 9d)
= 10a + 45d ...(2)
Step 3: Solve the system of equations
Now, we have two equations:
20 = a + 2d ...(1)
275 = 10a + 45d ...(2)
We can solve this system of equations to find the values of "a" and "d".
By rearranging equation (1), we get:
a = 20 - 2d
Substituting this into equation (2), we have:
275 = 10(20 - 2d) + 45d
= 200 - 20d + 45d
= 200 + 25d
25d = 275 - 200
25d = 75
d = 75 / 25
d = 3
Substituting the value of d back into equation (1), we can solve for "a":
20 = a + 2(3)
20 = a + 6
a = 20 - 6
a = 14
Therefore, the first term (a) of the linear sequence is 14, and the common difference (d) is 3.
a+2d = 20
10/2 (2a+19d) = 275
or, to make things easier,
a + 2d = 20
10a + 95d = 275
Now solve for a and d as usual