The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by $ \frac{21}{2} $. Then the number of terms which are integers in the A.P. is:
Show Hint
For arithmetic progressions, when the number of terms is even, you can use the sum of the odd and even terms to calculate the common difference and the first term.
The sum of the even-indexed terms is given by Equation 1:
\[
a_2 + a_4 + \cdots + a_{2n} = 30 \quad \text{(Equation 1)}
\]
The sum of the odd-indexed terms is given by Equation 2:
\[
a_1 + a_3 + \cdots + a_{2n-1} = 24 \quad \text{(Equation 2)}
\]
Subtracting Equation 2 from Equation 1 yields:
\[
(a_2 - a_1) + (a_4 - a_3) + \cdots + (a_{2n} - a_{2n-1}) = 6
\]
This simplifies to:
\[
\frac{n}{2} \cdot d = 6 \quad \text{(where \( d \) is the common difference)}
\]
From this, we get:
\[
nd = 6 \quad \Rightarrow \quad n = 12
\]
Using the equation \( a_{n} - a_1 = \frac{21}{2} \) and substituting \( n=12 \), we have:
\[
12d - d = \frac{21}{2} \quad \Rightarrow \quad 11d = \frac{21}{2}
\]
Solving for \( d \):
\[
d = \frac{3}{2}
\]
The sum of the odd terms can be calculated using the formula \( S_{\text{odd}} = \frac{4}{2} \left[ 2a_1 + (4-1) \cdot d \right] = 24 \).
Substituting the value of \( d \):
\[
2 \left[ 2a_1 + 3 \cdot \frac{3}{2} \right] = 24
\]
\[
2a_1 + \frac{9}{2} = 12
\]
\[
2a_1 = 12 - \frac{9}{2} = \frac{24-9}{2} = \frac{15}{2}
\]
\[
a_1 = \frac{15}{4}
\]
The arithmetic progression is: \( \frac{15}{4}, \frac{21}{4}, \frac{27}{4}, \frac{33}{4}, \dots \)
There seems to be a discrepancy in the provided A.P. \( \frac{3}{2}, \, 3, \, 9, \, 12, 15, 21, 9, 21, \dots \) and the calculated values. Assuming the calculations for \(n\) and \(d\) are correct based on the initial equations, the first term \(a_1\) should be \( \frac{15}{4} \). If \(a_1 = \frac{3}{2}\) and \(d=\frac{3}{2}\), then the sum of odd terms would be:
\[
S_{\text{odd}} = \frac{n}{2} [2a_1 + (n-1)d] = \frac{12}{2} [2(\frac{3}{2}) + (12-1)\frac{3}{2}] = 6 [3 + \frac{33}{2}] = 6 [\frac{6+33}{2}] = 6 [\frac{39}{2}] = 3 \times 39 = 117 eq 24
\]
There appears to be an inconsistency in the problem statement or the provided sequence.
If we assume the statement "The sum of odd terms: \(S_{\text{odd}} = \frac{4}{2} \left[ 2a_1 + (4-1) \cdot d \right] = 24\)" implies \(n=4\) in that specific calculation, and that the previous \(n=12\) and \(d=\frac{3}{2}\) are correct, then the \(a_1 = \frac{3}{2}\) calculation is derived from this \(n=4\) context, which contradicts \(n=12\).
Revisiting the sum of odd terms with \(n=12\) and \(d=\frac{3}{2}\) to match \(S_{\text{odd}} = 24\):
\[
S_{\text{odd}} = \frac{12}{2} [2a_1 + (12-1) \frac{3}{2}] = 6 [2a_1 + \frac{33}{2}] = 24
\]
\[
2a_1 + \frac{33}{2} = 4
\]
\[
2a_1 = 4 - \frac{33}{2} = \frac{8-33}{2} = -\frac{25}{2}
\]
\[
a_1 = -\frac{25}{4}
\]
With \(a_1 = -\frac{25}{4}\) and \(d=\frac{3}{2}\), the A.P. would be \( -\frac{25}{4}, -\frac{19}{4}, -\frac{13}{4}, \dots \)
The provided A.P. \( \frac{3}{2}, \, 3, \, 9, \, 12, 15, 21, 9, 21, \dots \) does not appear to be a standard arithmetic progression, as the differences between consecutive terms are not constant (e.g., \(3 - \frac{3}{2} = \frac{3}{2}\), \(9 - 3 = 6\)).
Assuming the calculation \(n=12\) and \(d=\frac{3}{2}\) is the primary derivation, and that \(a_1 = \frac{3}{2}\) was obtained from a separate calculation not fully detailed or consistent with the sum conditions.
The provided A.P. appears to be incorrect or does not follow the derived parameters.
The number of terms derived from the sum equations is 4.
