The second and sixth terms of an arithmetic sequence is 2:5 and their sum is 28. Find the sum of the first 10th term.
(a+d)/(a+5d) = 2/5
a+d + a+5d = 28
find a and d, and then
S10 = 10/2 (2a+9d)
To find the sum of the first 10 terms of an arithmetic sequence, we need to know the common difference (d).
First, let's find the common difference (d) using the given information. The second term is 5, and the sixth term is 28 - 5 = 23.
The formula to find the nth term of an arithmetic sequence is:
an = a1 + (n-1)d,
where an is the nth term, a1 is the first term, n is the position of the term, and d is the common difference.
Using the given information, we plug in the values:
5 = a1 + (2-1)d, ...(1)
23 = a1 + (6-1)d. ...(2)
Now we have a system of two equations (equation 1 and equation 2) with two variables (a1 and d). Let's solve these equations simultaneously.
From equation 1, we get:
a1 + d = 5.
From equation 2, we get:
a1 + 5d = 23.
To eliminate a1, we subtract equation 1 from equation 2:
(a1 + 5d) - (a1 + d) = 23 - 5.
4d = 18.
Dividing both sides by 4, we get:
d = 18/4 = 4.5.
Now that we have found the common difference (d) as 4.5, we can find the first term (a1) using equation 1:
a1 + d = 5.
a1 + 4.5 = 5.
a1 = 5 - 4.5 = 0.5.
Now that we have found a1 (first term) as 0.5 and d (common difference) as 4.5, we can find the sum of the first 10 terms using the formula for the sum of an arithmetic sequence:
Sn = (n/2)(2a1 + (n-1)d),
where Sn is the sum of n terms, n is the number of terms, a1 is the first term, and d is the common difference.
Plugging in the values, we have:
S10 = (10/2)(2(0.5) + (10-1)(4.5)).
S10 = 5(1 + 9(4.5)).
S10 = 5(1 + 40.5).
S10 = 5(41.5).
S10 = 207.5.
Therefore, the sum of the first 10 terms of the arithmetic sequence is 207.5.