the sum of 8th term in an a.p is 150 while the sum of 20th term is 880.find the 43rd term and the sum of the 12term
Just change the numbers in Reiny's solution here:
http://www.jiskha.com/display.cgi?id=1455402753
There are no solvings here
To find the 43rd term and the sum of the 12th term of an arithmetic progression (AP), we need to use the formulas for the nth term and the sum of n terms of an AP.
The nth term formula for an AP is given by:
\[a_n = a + (n-1)d\]
where:
- \(a\) is the first term of the AP,
- \(n\) is the position of the term we want to find,
- \(d\) is the common difference between consecutive terms.
The sum of the n terms formula for an AP is given by:
\[S_n = \frac{n}{2}[2a + (n-1)d]\]
where:
- \(S_n\) is the sum of the first n terms of the AP,
- \(a\) is the first term of the AP,
- \(n\) is the total number of terms in the AP,
- \(d\) is the common difference between consecutive terms.
Let's solve the problem step by step:
1. Find the common difference (d) using the given information about the 8th and 20th terms:
The sum of the 8th term is 150, so:
\[S_8 = \frac{8}{2}[2a + (8-1)d] = 150\]
Simplifying, we have:
\[4[2a + 7d] = 150\]
\[8a + 28d = 150 \tag{1}\]
Similarly, the sum of the 20th term is 880, so:
\[S_{20} = \frac{20}{2}[2a + (20-1)d] = 880\]
Simplifying, we have:
\[10[2a + 19d] = 880\]
\[20a + 190d = 880 \tag{2}\]
To solve equations (1) and (2), we can use any appropriate method, such as substitution or elimination. For the sake of brevity, let's solve them directly.
2. Solve equations (1) and (2) simultaneously to find the values of \(a\) and \(d\):
Multiplying equation (1) by 10 gives:
\[80a + 280d = 1500 \tag{3}\]
Subtracting equation (2) from equation (3) gives:
\[80a + 280d - (20a + 190d) = 1500 - 880\]
\[60a + 90d = 620 \tag{4}\]
Dividing equation (4) by 30 gives:
\[2a + 3d = \frac{31}{3} \tag{5}\]
Subtracting equation (5) from equation (1) (or equation (2)) gives:
\[(8a + 28d) - (2a + 3d) = 150 - \frac{31}{3}\]
\[6a + 25d = \frac{269}{3}\]
Simplifying, we get:
\[6a + 25d = \frac{269}{3} \tag{6}\]
We now have a system of two equations:
\[2a + 3d = \frac{31}{3} \tag{5}\]
\[6a + 25d = \frac{269}{3} \tag{6}\]
Solving this system of equations will give us the values of \(a\) and \(d\).