Find the partial sum Sn of the arithmetic sequence that satisfies the given conditions.

a2 = 5, a5 = 6.5, n = 15
S15 =

S15? Sounds like a distant cousin of R2-D2. But let's get back to the math problem. We have a2 = 5 and a5 = 6.5, which means we're dealing with an arithmetic sequence. To find S15, we need to find the common difference (d) first. Any clue on that?

Let's find the common difference first.

We know that a2 = a1 + d, where a1 is the first term and d is the common difference.
Given that a2 = 5, we can write the equation as: 5 = a1 + d. ----(1)

Similarly, a5 = a1 + 4d, where a5 is the fifth term.
Given that a5 = 6.5, we can write the equation as: 6.5 = a1 + 4d. ----(2)

To find the value of the common difference, we can subtract equation (1) from equation (2):

6.5 - 5 = (a1 + 4d) - (a1 + d)
1.5 = 3d
Dividing both sides by 3, we get:
d = 0.5

Now that we have the common difference, we can find the value of the first term, a1, by substituting d = 0.5 into equation (1):

5 = a1 + 0.5
Subtracting 0.5 from both sides, we get:
a1 = 4.5

To find the partial sum S15, we will use the formula:

Sn = (n/2)(2a1 + (n-1)d)

Substituting the values we found:
n = 15
a1 = 4.5
d = 0.5

Sn = (15/2)(2(4.5) + (15-1)(0.5))
Sn = 7.5(9 + 14(0.5))
Sn = 7.5(9 + 7)
Sn = 7.5(16)
Sn = 120

Therefore, S15 = 120.

To find the partial sum Sn of the arithmetic sequence, we first need to find the common difference (d) and the first term (a1) of the sequence.

We know that a2 = 5, which means the second term of the sequence is 5. The formula for the nth term of an arithmetic sequence is given by:
an = a1 + (n-1)d

Substituting n = 2 and a2 = 5 into the formula, we have:
5 = a1 + (2-1)d
5 = a1 + d

Similarly, we know that a5 = 6.5, which means the fifth term of the sequence is 6.5. Substituting n = 5 and a5 = 6.5 into the formula, we have:
6.5 = a1 + (5-1)d
6.5 = a1 + 4d

We now have a system of two equations:
1) 5 = a1 + d
2) 6.5 = a1 + 4d

We can solve this system of equations to find the values of a1 and d. Subtracting equation 1 from equation 2, we get:
6.5 - 5 = (a1 + 4d) - (a1 + d)
1.5 = 3d
d = 1.5/3
d = 0.5

Substituting the value of d into equation 1, we have:
5 = a1 + 0.5
a1 = 5 - 0.5
a1 = 4.5

Now that we have the values of a1 and d, we can find the partial sum Sn using the formula:
Sn = (n/2)(a1 + an)

Substituting n = 15, a1 = 4.5, and d = 0.5 into the formula, we have:
S15 = (15/2)(4.5 + (4.5 + (15-1)0.5))
S15 = (15/2)(4.5 + (4.5 + 14(0.5)))
S15 = (15/2)(4.5 + (4.5 + 7))
S15 = (15/2)(4.5 + 11.5)
S15 = (15/2)(16)
S15 = 15 * 8
S15 = 120

Therefore, the partial sum S15 of the arithmetic sequence is 120.

using the standard notation of a and d,

a = 5
a+4d = 6.5
then 5 + d = 6.5
d = 1.5

sum(15) = (15/2)(2(5) + 14(1.5))
= 232.5