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The question you've provided is related to an Arithmetic Progression (A.P.):

The A.P. given is:  
20, 17, 14, ...

The sum of a certain number of terms of this A.P. is 65. We are asked to find:
- (a) The number of terms.
- (b) The last term.

### Step 1: Formula for Sum of an Arithmetic Progression
The formula for the sum \( S_n \) of the first \( n \) terms of an arithmetic progression is:
\[
S_n = \frac{n}{2} \times (2a + (n - 1) d)
\]
Where:
- \( a \) is the first term.
- \( d \) is the common difference.
- \( n \) is the number of terms.
- \( S_n \) is the sum of the first \( n \) terms.

### Step 2: Identifying Given Values
From the sequence: 20, 17, 14, ..., we have:
- \( a = 20 \) (first term)
- \( d = 17 - 20 = -3 \) (common difference)
- \( S_n = 65 \) (the given sum)

We need to find the number of terms \( n \).

### Step 3: Substituting Values into the Formula
Substitute the known values into the sum formula:
\[
65 = \frac{n}{2} \times (2 \times 20 + (n - 1) \times (-3))
\]
\[
65 = \frac{n}{2} \times (40 - 3n + 3)
\]
\[
65 = \frac{n}{2} \times (43 - 3n)
\]
\[
130 = n \times (43 - 3n)
\]
\[
130 = 43n - 3n^2
\]

### Step 4: Solving the Quadratic Equation
Rearrange the equation:
\[
3n^2 - 43n + 130 = 0
\]

Use the quadratic formula:
\[
n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where \( a = 3 \), \( b = -43 \), and \( c = 130 \).

\[
n = \frac{43 \pm \sqrt{(-43)^2 - 4(3)(130)}}{2(3)}
\]
\[
n = \frac{43 \pm \sqrt{1849 - 1560}}{6}
\]
\[
n = \frac{43 \pm \sqrt{289}}{6}
\]
\[
n = \frac{43 \pm 17}{6}
\]

This gives two solutions:
\[
n = \frac{43 + 17}{6} = \frac{60}{6} = 10
\]
or
\[
n = \frac{43 - 17}{6} = \frac{26}{6} = 4.33 \quad (\text{which is not possible since \( n \) must be an integer})
\]

Thus, \( n = 10 \).

### Step 5: Finding the Last Term
The last term \( l_n \) of an arithmetic progression is given by:
\[
l_n = a + (n - 1)d
\]
Substitute \( n = 10 \), \( a = 20 \), and \( d = -3 \):
\[
l_{10} = 20 + (10 - 1) \times (-3)
\]
\[
l_{10} = 20 + 9 \times (-3) = 20 - 27 = -7
\]

### Final Answers:
- (a) The number of terms is \( 10 \).
- (b) The last term is \( -7 \).

Would you like further details or clarification? Here are five related questions to consider:

1. How do we derive the formula for the sum of an arithmetic progression?
2. What are other types of sequences that use similar methods?
3. How can you solve for the number of terms in a geometric progression?
4. Why is the quadratic formula necessary when dealing with A.P. sums?
5. What are some real-world applications of arithmetic progressions?

**Tip:** Always ensure that the common difference is correctly identified, as it directly affects the sum and general term of the A.P.

The sum of a certain number of terms of the Arithmetic Progression (A.P.) 20, 17, 14, ... is 65. Find the: (a) number of terms, (b) last term.

Learn how to solve a problem involving the sum of terms in an arithmetic progression (A.P.) with first term 20 and common difference -3. Detailed step-by-step solution covers finding both the number of terms and the last term. Suitable for students in Grades 8-10.

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