The su吗the sum of the 1st twenty-one of linear sequence is 28, and sum of the first twenty -eight terms is 21.find (a) the term of the sequence that is equal to 0. (b) the sum of the terms preceding the term 0
sum(21) = 28
(21/2)(2a + 20d) = 28
2a+20d = 56/21
sum(28) = 21
(28/2)(2a + 27d) = 21
2a+27d = 42/28 = 3/2
subtract the two equations:
7d = 3/2 - 56/21 = -7/6
d = -1/6
in 2a + 20d = 56/21
2a - 20/6 = 56/21
2a = 6
a = 3
so we need:
a + (n-1)d = 0
3 + (n-1)(-1/6) = 0
3 - (-1/6)n + 1/6 = 0
19/6 = n/6
n = 19
sum(19) = 0
check:
a+18d
= 3 + 18(-1/6) = 0
b)
so sum(18) = ......... (your turn)
S18 =18÷2(2a+[n-1]d).
S18=9(6+[17]-1÷6
S18=9(6-17÷6)
S18=28.53
To solve this problem, we can use the formulas for the sum of the terms of an arithmetic sequence.
(a) To find the term of the sequence that is equal to 0, we can set up an equation using the sum of the first twenty-one terms. Let's call the first term of the sequence "a" and the common difference "d".
The sum of the first twenty-one terms can be calculated using the formula:
S21 = (21/2)(2a + (21-1)d) = 28
Simplifying the equation:
21a + 210d - 231d = 56
21a - 21d = 56
a - d = 56/21
a - d = 8/3
Similarly, using the sum of the first twenty-eight terms, we can set up another equation:
S28 = (28/2)(2a + (28-1)d) = 21
28a + 378d - 406d = 42
Simplifying the equation:
28a - 28d = 42
a - d = 42/28
a - d = 3/2
Now, we can solve the system of equations:
a - d = 8/3
a - d = 3/2
Subtracting the second equation from the first, we get:
(8/3) - (3/2) = (16/6) - (9/6)
(8/3) - (3/2) = 7/6
Therefore, the term of the sequence that is equal to 0 is the 7/6th term.
(b) To find the sum of the terms preceding the term 0, we can use the formula for the sum of an arithmetic series:
Sn = (n/2)(2a + (n-1)d)
Substituting the known values:
S(n-1) = ((7/6)-1)/2)*(2a + ((7/6)-2)*d)
Since we know a - d = 8/3, we can substitute this value into the equation:
S(n-1) = ((7/6)-1)/2)*(2(a-d) + ((7/6)-2)*d)
Simplifying the equation:
S(n-1) = ((7/6)-1)/2)*(2(8/3) + ((7/6)-2)*d)
Now, we can calculate the sum of the terms preceding the term 0 by substituting the values a = 8/3 and d into the equation and solving for S(n-1).
To find the answers to both parts (a) and (b) of the question, we need to analyze the given information and apply the formula to find the sum of an arithmetic sequence.
(a) Finding the term of the sequence that is equal to 0:
Let's assume that the first term of the arithmetic sequence is 'a,' and the common difference is 'd.' The sum of the first twenty-one terms can be calculated using the formula:
Sum = (n/2) * (2a + (n-1)d)
Given that the sum of the first twenty-one terms is 28, we can substitute this information into the formula and solve for 'a':
28 = (21/2) * (2a + (21-1)d)
28 = 10.5 * (2a + 20d)
Simplifying the equation, we get:
28 = 21a + 210d ---(1)
Similarly, we can find the sum of the first twenty-eight terms using the same formula:
21 = (28/2) * (2a + (28-1)d)
21 = 14 * (2a + 27d)
Simplifying this equation, we get:
21 = 28a + 378d ---(2)
Now, we can solve equations (1) and (2) simultaneously to find the values of 'a' and 'd':
Subtracting equation (2) from equation (1), we get:
28 - 21 = (21a + 210d) - (28a + 378d)
7 = -7a - 168d
Rearranging this equation, we have:
7a + 168d = -7 ---(3)
Simplifying equation (3), we can rewrite it as:
a + 24d = -1
Since we have two variables, 'a' and 'd,' and one equation, we need another equation to solve for their values. We can use the second piece of information given, which is the sum of the first twenty-eight terms being 21.
Substituting this into the formula, we get an additional equation:
21 = (28/2) * (2a + (28-1)d)
21 = 14 * (2a + 27d)
Simplifying equation (4), we have:
21 = 28a + 378d ---(4)
Now, we have a system of equations (3) and (4) that we can solve simultaneously.
Multiplying equation (3) by 28 and equation (4) by 7, we get:
196a + 4704d = -196 ---(5)
196a + 2646d = 147 ---(6)
Subtracting equation (6) from equation (5), we can eliminate the variable 'a':
4704d - 2646d = -196 - 147
2058d = -343
Simplifying equation (7), we find:
d = -343/2058
d = -1/6
Substituting this value for 'd' into equation (3), we can solve for 'a':
a + 24(-1/6) = -1
a - 4 = -1
a = 3
Now that we have the values of 'a' and 'd,' we can determine the term of the sequence that is equal to 0.
The nth term of an arithmetic sequence can be found using the formula:
an = a + (n-1)d
Substituting 'a' and 'd' into this formula, we get:
an = 3 + (n-1)(-1/6)
an = 3 - (n-1)/6
To find the term that is equal to 0 (an = 0), we can solve the equation:
0 = 3 - (n-1)/6
Rearranging the equation, we have:
(n-1)/6 = 3
n-1 = 18
n = 19
Therefore, the term of the sequence that is equal to 0 is the 19th term.
(b) Finding the sum of the terms preceding the term 0:
To find the sum of the terms preceding the term 0, we can calculate the sum of the first 18 terms (before reaching 19) using the sum formula for an arithmetic sequence.
The sum of the first 'n' terms can be calculated using the formula:
Sum = (n/2) * (2a + (n-1)d)
Substituting the known values, we can find the sum of the first 18 terms:
Sum = (18/2) * (2a + (18-1)d)
Sum = 9 * (2a + 17d)
Using the values we previously calculated for 'a' and 'd':
Sum = 9 * (2(3) + 17(-1/6))
Sum = 9 * (6 - 17/6)
Sum = 9 * (36/6 - 17/6)
Sum = 9 * 19/6
Sum = 57/2
Therefore, the sum of the terms preceding the term 0 is 57/2.