To solve this problem, we need to find the first term ([tex]\(a\)[/tex]) and the common difference ([tex]\(d\)[/tex]) of the arithmetic progression (A.P.).
Given:
1. Sum of the first three terms ([tex]\(S_3\)[/tex]) is 48.
2. Sum of the first five terms ([tex]\(S_5\)[/tex]) is 110.
3. The twelfth term ([tex]\(T_{12}\)[/tex]) is 30.
First, we'll set up the equations using these conditions.
### Step 1: Setting up equations
The sum of the first [tex]\(n\)[/tex] terms of an A.P. can be given by:
[tex]\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \][/tex]
1. Using [tex]\(S_3 = 48\)[/tex],
[tex]\[ S_3 = \frac{3}{2} \times (2a + 2d) = 48 \][/tex]
[tex]\[ 3(a + d) = 48 \][/tex]
[tex]\[ a + d = 16 \quad \text{(Equation 1)} \][/tex]
2. Using [tex]\(S_5 = 110\)[/tex],
[tex]\[ S_5 = \frac{5}{2} \times (2a + 4d) = 110 \][/tex]
[tex]\[ 5(a + 2d) = 110 \][/tex]
[tex]\[ a + 2d = 22 \quad \text{(Equation 2)} \][/tex]
### Step 2: Solving for [tex]\(a\)[/tex] and [tex]\(d\)[/tex]
We can solve the system of linear equations (Equation 1 and Equation 2):
From Equation 1,
[tex]\[ a + d = 16 \][/tex]
From Equation 2,
[tex]\[ a + 2d = 22 \][/tex]
Subtract Equation 1 from Equation 2,
[tex]\[ (a + 2d) - (a + d) = 22 - 16 \][/tex]
[tex]\[ d = 6 \][/tex]
Substituting [tex]\(d = 6\)[/tex] in Equation 1,
[tex]\[ a + 6 = 16 \][/tex]
[tex]\[ a = 10 \][/tex]
So, we have:
[tex]\[ a = 10 \][/tex]
[tex]\[ d = 6 \][/tex]
### Step 3: Finding the 15th term [tex]\(T_{15}\)[/tex]
The [tex]\(n\)[/tex]-th term of an A.P. is given by:
[tex]\[ T_n = a + (n - 1)d \][/tex]
For the 15th term:
[tex]\[ T_{15} = a + 14d \][/tex]
[tex]\[ T_{15} = 10 + 14 \cdot 6 \][/tex]
[tex]\[ T_{15} = 10 + 84 \][/tex]
[tex]\[ T_{15} = 94 \][/tex]
### Step 4: Finding the sum of the first 15 terms [tex]\(S_{15}\)[/tex]
Using the sum formula:
[tex]\[ S_{15} = \frac{15}{2} \times (2a + 14d) \][/tex]
[tex]\[ S_{15} = \frac{15}{2} \times (2 \cdot 10 + 14 \cdot 6) \][/tex]
[tex]\[ S_{15} = \frac{15}{2} \times (20 + 84) \][/tex]
[tex]\[ S_{15} = \frac{15}{2} \times 104 \][/tex]
[tex]\[ S_{15} = 15 \times 52 \][/tex]
[tex]\[ S_{15} = 780 \][/tex]
### Results
- The [tex]\(15^{\text{th}}\)[/tex] term of the A.P. is [tex]\(94\)[/tex].
- The sum of the first 15 terms of the A.P. is [tex]\(780\)[/tex].
These are the step-by-step solutions to find the required terms of the arithmetic progression.
