In an arithmetic sequence , the 9th term is twice the 3rd term and 15th term is 80. Find the common difference and the sum of the terms from from 9th to 15th is inclusive
Just change your English to math, using the definitions you learned.
"the 9th term is twice the 3rd term" ---> a + 8d = 2(a + 2d)
a + 8d = 2a + 4d
a = 4d
"15th term is 80" ----> a + 14d = 80
4d + 14d = 80
take it from there.
Use your sum of terms formula for the last part of your question.
Take the sum of 15 terms and subtract the sum of 8 terms from that.
To find the common difference of an arithmetic sequence, we can use the formula:
a_n = a_1 + (n - 1)d
Where:
a_n represents the nth term,
a_1 represents the first term,
n represents the term number,
and d represents the common difference.
Given that the 9th term is twice the 3rd term, we can write the equation as:
a_9 = 2 * a_3
Substituting the formula for both terms, we get:
a_1 + 8d = 2(a_1 + 2d)
Simplifying the equation, we have:
a_1 + 8d = 2a_1 + 4d
Subtracting a_1 and 4d from both sides, we get:
4d = a_1
This equation tells us that the common difference is one-fourth of the first term.
Now, let's find the value of the common difference.
Given that the 15th term is 80, we can write the equation as:
a_15 = a_1 + 14d = 80
Substituting a_1 = 4d, we get:
4d + 14d = 80
Combining like terms, we have:
18d = 80
Dividing both sides by 18, we find:
d ≈ 4.44
Therefore, the common difference is approximately 4.44.
Next, let's find the sum of the terms from the 9th to the 15th term inclusive.
We can use the formula for the sum of an arithmetic series:
S_n = (n/2)(a_1 + a_n)
Substituting the values, we get:
S_7 = (7/2)(a_1 + a_15)
Since we know a_1 = 4d and a_15 = 80, we can substitute these values in the formula:
S_7 = (7/2)(4d + 80)
Substituting the value of d ≈ 4.44, we find:
S_7 = (7/2)(4 * 4.44 + 80) = (7/2)(17.76 + 80) = (7/2)(97.76) = 342.16
Therefore, the sum of the terms from the 9th to the 15th term inclusive is approximately 342.16.
To find the common difference in an arithmetic sequence, we need to use the formula:
nth term = a + (n - 1)d,
where "a" represents the first term, "n" represents the position of the term, and "d" represents the common difference.
Using the given information, we have two equations:
9th term = 2 * 3rd term,
15th term = 80.
Let's solve for the common difference and find the value of "d" first.
From the first equation, we have:
9th term = 2 * 3rd term.
a + 8d = 2(a + 2d).
Simplifying this equation, we get:
a + 8d = 2a + 4d,
8d - 4d = 2a - a,
4d = a.
Substituting this into the second equation:
15th term = 80,
a + 14d = 80.
Substituting "4d" for "a," we have:
4d + 14d = 80,
18d = 80.
Divide both sides by 18:
d = 80/18 = 4.44 (rounded to two decimal places).
Therefore, the common difference, "d," in the arithmetic sequence is approximately 4.44.
To find the sum of the terms from the 9th to the 15th term, inclusive, we can use the formula for the sum of an arithmetic series:
Sum = (n/2)(2a + (n - 1)d),
where "n" represents the number of terms and "a" represents the first term.
Substituting the known values, we have:
n = 15 - 9 + 1 = 7 (since we are including the 9th and 15th terms),
a = 9th term = a + 8d = 4d + 8d = 12d.
Substituting these values into the sum formula:
Sum = (7/2)(2(12d) + (7 - 1)d),
Sum = (7/2)(24d + 6d),
Sum = (7/2)(30d),
Sum = 105d.
Substituting the value of "d" (4.44) we found earlier:
Sum = 105 * 4.44 = 466.20.
Therefore, the sum of the terms from the 9th to the 15th term, inclusive, is approximately 466.20.
a + r (n-1)
a + 8 r = 2(a+2r) = 2 a +4 r
a + 14 r = 80
a + 14 r = 80
a - 4 r = 0
------------------- suubtract
18 r = 80
r = 4.44
a = 4 r = 17.78