Problem:
If the sum of a certain number of terms, starting from first, of an A.P. 25, 22, 19, ..., is 116, find the last term taken in the sum.
Solution:
#### Step 1: Identify the given information
- The arithmetic progression (A.P.) is: \(25, 22, 19, \ldots\).
- The first term (\(a\)) is: \(25\).
- The common difference (\(d\)) is: \(22 - 25 = -3\).
- The sum of the first \(n\) terms (\(S_n\)) is given as: \(116\).
We need to find the last term (\(l\)) of the sequence.
#### Step 2: Use the formula for the sum of the first \(n\) terms of an A.P.
The sum of the first \(n\) terms of an A.P. is given by:
\[
S_n = \frac{n}{2} \left[2a + (n-1)d\right]
\]
Substitute the known values:
\[
116 = \frac{n}{2} \left[2(25) + (n-1)(-3)\right]
\]
Simplify inside the brackets:
\[
116 = \frac{n}{2} \left[50 - 3(n-1)\right]
\]
\[
116 = \frac{n}{2} \left[50 - 3n + 3\right]
\]
\[
116 = \frac{n}{2} \left[53 - 3n\right]
\]
#### Step 3: Solve for \(n\)
Multiply both sides by 2 to eliminate the fraction:
\[
232 = n(53 - 3n)
\]
Expand and rearrange into standard quadratic form:
\[
232 = 53n - 3n^2
\]
\[
3n^2 - 53n + 232 = 0
\]
#### Step 4: Solve the quadratic equation
Use the quadratic formula:
\[
n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \(a = 3\), \(b = -53\), and \(c = 232\). Substitute these values:
\[
n = \frac{-(-53) \pm \sqrt{(-53)^2 - 4(3)(232)}}{2(3)}
\]
\[
n = \frac{53 \pm \sqrt{2809 - 2784}}{6}
\]
\[
n = \frac{53 \pm \sqrt{25}}{6}
\]
\[
n = \frac{53 \pm 5}{6}
\]
This gives two possible solutions:
\[
n = \frac{53 + 5}{6} = \frac{58}{6} = \frac{29}{3} \quad \text{(not an integer, discard)}
\]
\[
n = \frac{53 - 5}{6} = \frac{48}{6} = 8
\]
So, \(n = 8\).
#### Step 5: Find the last term (\(l\))
The last term (\(l\)) of an A.P. is given by:
\[
l = a + (n-1)d
\]
Substitute \(a = 25\), \(d = -3\), and \(n = 8\):
\[
l = 25 + (8-1)(-3)
\]
\[
l = 25 + 7(-3)
\]
\[
l = 25 - 21
\]
\[
l = 4
\]
Final Answer:
\[
\boxed{4}
\]
Parent Tip: Review the logic above to help your child master the concept of series and sequences worksheet.
