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Arithmetic Progressions
EXERCISE 5.2
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

Find the number of terms in each of the following APs : (i) 7, 13, 19, . . . , 205 (ii) 18, 1 15 2 , 13, . . . , – 47

Step-by-Step Solution

Key Concept: For an arithmetic progression (AP) with first term \(a\), common difference \(d\) and last term \(l\), the number of terms \(n\) is given by the formula \[ n = \frac{l - a}{d} + 1 \] provided \(d <br>eq 0\). The common difference is obtained from any two successive terms: \(d = a_{2} - a_{1}\).
### (i) \(7,\;13,\;19,\;\dotsc,\;205\)
1. Identify the first term \(a\) and the common difference \(d\).
- \(a = 7\)
- \(d = 13 - 7 = 6\)
2. Use the formula for the number of terms:
\[ n = \frac{l - a}{d} + 1 \]
where \(l = 205\).
3. Substitute the values:
\[ n = \frac{205 - 7}{6} + 1 = \frac{198}{6} + 1 = 33 + 1 = 34 \]
4. Hence, the AP contains 34 terms.

### (ii) \(18,\;\frac{15}{2},\;13,\;\dotsc,\;-47\)
1. First term \(a = 18\).
2. The common difference \(d\) can be obtained from the first and third terms (since the second term given is ambiguous, we use the first and third terms which are definitely part of the AP):
\[ d = \frac{13 - 18}{2} = -\frac{5}{2} \]
3. Apply the formula for the number of terms with \(l = -47\):
\[ n = \frac{-47 - 18}{-\frac{5}{2}} + 1 \]
4. Simplify the numerator:
\[ -47 - 18 = -65 \]
5. Divide by \(-\frac{5}{2}\):
\[ \frac{-65}{-\frac{5}{2}} = -65 \times \left(-\frac{2}{5}\right) = \frac{130}{5} = 26 \]
6. Add 1:
\[ n = 26 + 1 = 27 \]
7. Therefore, the AP contains 27 terms.

Note: The second AP in the question has a typographical inconsistency in the second term. Using the first and third terms (which are clearly part of the progression) yields a consistent common difference and the above result.

Correct Answer: (i) 34 terms (ii) 27 terms
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