Answer :
To solve this problem, we need to use formulas and properties of Arithmetic Progressions (A.P.). Let's break it down step-by-step:
### Given Information:
1. The sum of the first 15 terms [tex]\( S_{15} \)[/tex] is 360
2. The sum of the first 19 terms [tex]\( S_{19} \)[/tex] is 570
3. The sixth term of the A.P. is 5 times the first term.
We need to find the number of terms [tex]\( n \)[/tex] such that the sum of the first [tex]\( n \)[/tex] terms is 759.
### Steps to Solve:
1. Sum of an Arithmetic Progression:
The formula for the sum [tex]\( S_n \)[/tex] of the first [tex]\( n \)[/tex] terms of an A.P. is given by:
[tex]\[ S_n = \frac{n}{2} \left( 2a + (n-1)d \right) \][/tex]
where [tex]\( a \)[/tex] is the first term and [tex]\( d \)[/tex] is the common difference.
2. Using the given sums [tex]\( S_{15} \)[/tex] and [tex]\( S_{19} \)[/tex]:
For the first 15 terms:
[tex]\[ S_{15} = \frac{15}{2} \left( 2a + 14d \right) = 360 \][/tex]
Simplifying:
[tex]\[ 15 \left( 2a + 14d \right) = 720 \implies 2a + 14d = 48 \implies a + 7d = 24 \quad \text{(Equation 1)} \][/tex]
For the first 19 terms:
[tex]\[ S_{19} = \frac{19}{2} \left( 2a + 18d \right) = 570 \][/tex]
Simplifying:
[tex]\[ 19 \left( 2a + 18d \right) = 1140 \implies 2a + 18d = 60 \implies a + 9d = 30 \quad \text{(Equation 2)} \][/tex]
3. Solving the simultaneous equations (Equation 1 and Equation 2):
[tex]\[ a + 7d = 24 \quad \text{(Equation 1)} \][/tex]
[tex]\[ a + 9d = 30 \quad \text{(Equation 2)} \][/tex]
Subtract Equation 1 from Equation 2:
[tex]\[ (a + 9d) - (a + 7d) = 30 - 24 \][/tex]
[tex]\[ 2d = 6 \implies d = 3 \][/tex]
Substitute [tex]\( d = 3 \)[/tex] into Equation 1:
[tex]\[ a + 7 \cdot 3 = 24 \implies a + 21 = 24 \implies a = 3 \][/tex]
4. Finding the number of terms [tex]\( n \)[/tex] such that [tex]\( S_n = 759 \)[/tex]:
Using the sum formula again:
[tex]\[ S_n = \frac{n}{2} \left( 2a + (n-1)d \right) = 759 \][/tex]
Substitute [tex]\( a = 3 \)[/tex] and [tex]\( d = 3 \)[/tex]:
[tex]\[ 759 = \frac{n}{2} \left( 2 \cdot 3 + (n-1) \cdot 3 \right) \][/tex]
Simplify:
[tex]\[ 759 = \frac{n}{2} \left( 6 + 3n - 3 \right) \][/tex]
[tex]\[ 759 = \frac{n}{2} \left( 3 + 3n \right) \][/tex]
[tex]\[ 759 = \frac{3n(1 + n)}{2} \][/tex]
[tex]\[ 1518 = 3n(n + 1) \][/tex]
[tex]\[ 506 = n(n + 1) \][/tex]
[tex]\[ n^2 + n - 506 = 0 \][/tex]
Solving the quadratic equation [tex]\( n^2 + n - 506 = 0 \)[/tex]:
[tex]\[ n = \frac{-1 \pm \sqrt{1 + 4 \cdot 506}}{2} \][/tex]
[tex]\[ n = \frac{-1 \pm \sqrt{1 + 2024}}{2} \][/tex]
[tex]\[ n = \frac{-1 \pm \sqrt{2025}}{2} \][/tex]
[tex]\[ n = \frac{-1 \pm 45}{2} \][/tex]
Choosing the positive root:
[tex]\[ n = \frac{44}{2} = 22 \][/tex]
### Conclusion:
The number of terms from the first that give the sum 759 is [tex]\( 22 \)[/tex].
### Given Information:
1. The sum of the first 15 terms [tex]\( S_{15} \)[/tex] is 360
2. The sum of the first 19 terms [tex]\( S_{19} \)[/tex] is 570
3. The sixth term of the A.P. is 5 times the first term.
We need to find the number of terms [tex]\( n \)[/tex] such that the sum of the first [tex]\( n \)[/tex] terms is 759.
### Steps to Solve:
1. Sum of an Arithmetic Progression:
The formula for the sum [tex]\( S_n \)[/tex] of the first [tex]\( n \)[/tex] terms of an A.P. is given by:
[tex]\[ S_n = \frac{n}{2} \left( 2a + (n-1)d \right) \][/tex]
where [tex]\( a \)[/tex] is the first term and [tex]\( d \)[/tex] is the common difference.
2. Using the given sums [tex]\( S_{15} \)[/tex] and [tex]\( S_{19} \)[/tex]:
For the first 15 terms:
[tex]\[ S_{15} = \frac{15}{2} \left( 2a + 14d \right) = 360 \][/tex]
Simplifying:
[tex]\[ 15 \left( 2a + 14d \right) = 720 \implies 2a + 14d = 48 \implies a + 7d = 24 \quad \text{(Equation 1)} \][/tex]
For the first 19 terms:
[tex]\[ S_{19} = \frac{19}{2} \left( 2a + 18d \right) = 570 \][/tex]
Simplifying:
[tex]\[ 19 \left( 2a + 18d \right) = 1140 \implies 2a + 18d = 60 \implies a + 9d = 30 \quad \text{(Equation 2)} \][/tex]
3. Solving the simultaneous equations (Equation 1 and Equation 2):
[tex]\[ a + 7d = 24 \quad \text{(Equation 1)} \][/tex]
[tex]\[ a + 9d = 30 \quad \text{(Equation 2)} \][/tex]
Subtract Equation 1 from Equation 2:
[tex]\[ (a + 9d) - (a + 7d) = 30 - 24 \][/tex]
[tex]\[ 2d = 6 \implies d = 3 \][/tex]
Substitute [tex]\( d = 3 \)[/tex] into Equation 1:
[tex]\[ a + 7 \cdot 3 = 24 \implies a + 21 = 24 \implies a = 3 \][/tex]
4. Finding the number of terms [tex]\( n \)[/tex] such that [tex]\( S_n = 759 \)[/tex]:
Using the sum formula again:
[tex]\[ S_n = \frac{n}{2} \left( 2a + (n-1)d \right) = 759 \][/tex]
Substitute [tex]\( a = 3 \)[/tex] and [tex]\( d = 3 \)[/tex]:
[tex]\[ 759 = \frac{n}{2} \left( 2 \cdot 3 + (n-1) \cdot 3 \right) \][/tex]
Simplify:
[tex]\[ 759 = \frac{n}{2} \left( 6 + 3n - 3 \right) \][/tex]
[tex]\[ 759 = \frac{n}{2} \left( 3 + 3n \right) \][/tex]
[tex]\[ 759 = \frac{3n(1 + n)}{2} \][/tex]
[tex]\[ 1518 = 3n(n + 1) \][/tex]
[tex]\[ 506 = n(n + 1) \][/tex]
[tex]\[ n^2 + n - 506 = 0 \][/tex]
Solving the quadratic equation [tex]\( n^2 + n - 506 = 0 \)[/tex]:
[tex]\[ n = \frac{-1 \pm \sqrt{1 + 4 \cdot 506}}{2} \][/tex]
[tex]\[ n = \frac{-1 \pm \sqrt{1 + 2024}}{2} \][/tex]
[tex]\[ n = \frac{-1 \pm \sqrt{2025}}{2} \][/tex]
[tex]\[ n = \frac{-1 \pm 45}{2} \][/tex]
Choosing the positive root:
[tex]\[ n = \frac{44}{2} = 22 \][/tex]
### Conclusion:
The number of terms from the first that give the sum 759 is [tex]\( 22 \)[/tex].
