The sum of all five terms of an arithmetic sequence is

55. Find the sum of the first three terms of the sequence if the sum of the last three is 48.

S5 = 5/2 (2a+4d)

S2 = 2/2 (2a+d)
The sum of the last 3 is S5-S2, so we have
S2 = 55-48 = 7 = 2a+d

2a+d = 7
5/2 (2a+4d) = 55

a=1, d=5

So, S3 = 3/2 (2+2*5) = 18

Well, I guess this arithmetic sequence has a case of the "sum blues". Let's see if we can cheer it up with a little math humor!

Let's call the first term "a" and the common difference "d". The sum of all five terms is 55, so we can form the equation:

a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 55

Now let's address the other complaint, the sum of the last three terms being 48. We can write another equation for this:

(a + 2d) + (a + 3d) + (a+ 4d) = 48

Clown Bot has all the equations! Now it's time to solve them and make everyone happy.

Simplifying both equations, we get:

5a + 10d = 55 ----(1)
3a + 9d = 48 ----(2)

Now, let's put on our serious math hats for a moment and solve these equations together.

From equation (2), we can rearrange it to solve for a:

3a = 48 - 9d
a = 16 - 3d

Now substitute this value of a into equation (1):

5(16 - 3d) + 10d = 55
80 - 15d + 10d = 55
-5d = -25
d = 5

Now, substitute the value of d back into equation (2):

3a + 9(5) = 48
3a + 45 = 48
3a = 3
a = 1

So, the first term is 1 and the common difference is 5. Let's add some clown magic to find the sum of the first three terms of this sequence.

The first three terms are: 1, 6, and 11. Adding them up, we get:

1 + 6 + 11 = 18

Ta-da! The sum of the first three terms of the sequence is 18. Clown Bot saves the day once again!

Let's denote the first term of the arithmetic sequence as 'a', and the common difference as 'd'.

The sum of all five terms of the arithmetic sequence is given as 55, so we can express it as:

a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 55

Now, let's find the sum of the last three terms of the sequence, which is given as 48:

(a + 2d) + (a + 3d) + (a + 4d) = 48

We can solve these equations simultaneously to find the values of 'a' and 'd'.

From the first equation, we can simplify it as:

5a + 10d = 55 [equation 1]

From the second equation, we can simplify it as:

3a + 9d = 48 [equation 2]

To make calculations easier, we can divide both equations by their respective coefficients:

equation 1 / 5: a + 2d = 11 [equation 3]
equation 2 / 3: a + 3d = 16 [equation 4]

Now, we can solve equations 3 and 4 simultaneously:

Subtracting equation 3 from equation 4:
(a + 3d) - (a + 2d) = 16 - 11
a - a + 3d - 2d = 5d = 5

So, the value of d is 1.

Substituting the value of d = 1 in equation 3:
a + 2(1) = 11
a + 2 = 11
a = 11 - 2
a = 9

Therefore, we have found that the first term 'a' is 9, and the common difference 'd' is 1.

To find the sum of the first three terms of the sequence, we can use the formula:

Sum of n terms = (n/2) * [2a + (n-1)d]

Substituting n = 3, a = 9, and d = 1:
Sum of the first three terms = (3/2) * [2(9) + (3-1)(1)]
= (3/2) * [18 + 2]
= (3/2) * 20
= 30

So, the sum of the first three terms of the sequence is 30.

To solve this problem, we need to use the formulas for the sum of an arithmetic sequence.

Let's assume that the first term of the sequence is 'a' and the common difference is 'd'.

The sum of all five terms can be calculated using the formula:

Sum = (n/2) * (2a + (n-1)d)

Here, 'n' represents the number of terms in the sequence. In this case, we have five terms, so n = 5.

Given that the sum of all five terms is 55, we can substitute the values into the formula:

55 = (5/2) * (2a + 4d)
110 = 10a + 20d
Equation 1: 10a + 20d - 110 = 0

Now, let's find the sum of the last three terms, using the same formula:

Sum = (n/2) * (2a + (n-1)d)

Substituting n = 3, we get:

48 = (3/2) * (2a + 2d)
96 = 6a + 6d
Equation 2: 6a + 6d - 96 = 0

To solve these two equations simultaneously, we can use the method of substitution or elimination.

Let's divide equation 2 by 6:

a + d - 16 = 0
Equation 3: a + d = 16

Now, we can substitute this value of (a + d) in equation 1:

10(16) + 20d - 110 = 0
160 + 20d - 110 = 0
20d = -50
d = -50/20
d = -5/2

Now, substitute the value of 'd' back into Equation 3:

a + (-5/2) = 16
a - 5/2 = 16
a = 16 + 5/2
a = 16 + 2.5
a = 18.5

So, the first term 'a' is 18.5 and the common difference 'd' is -5/2.

To find the sum of the first three terms (a1 + a2 + a3), we can plug in the values into the formula:

Sum = 3/2 * (2a + (3-1)d)
Sum = (3/2) * (2(18.5) + 2*(-5/2))
Sum = (3/2) * (37 - 5)
Sum = (3/2) * 32
Sum = 48

Therefore, the sum of the first three terms of the arithmetic sequence is 48.