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प्रश्न
The number of terms of an A.P. is even; the sum of odd terms is 24, of the even terms is 30, and the last term exceeds the first by \[10 \frac{1}{2}\] , find the number of terms and the series.
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उत्तर
Let total number of terms be 2n.
According to question, we have:
\[a_1 + a_3 + . . . + a_{2n - 1} = 24 . . . (1)\]
\[ a_2 + a_4 + . . . + a_{2n} = 30 . . . (2)\]
\[\text { Subtracting (1) from (2), we get: } \]
\[\left( d + d + . . . + \text { upto n terms } \right) = 6\]
\[ \Rightarrow nd = 6 . . . (3)\]
\[\text { Given }: \]
\[ a_{2n} = a_1 + \frac{21}{2}\]
\[ \Rightarrow a_{2n} - a_1 = \frac{21}{2}\]
\[ \Rightarrow a + (2n - 1)d - a = \frac{21}{2} [ \because a_{2n} = a + (2n - 1)d, a_1 = a]\]
\[ \Rightarrow 2nd - d = \frac{21}{2}\]
\[ \Rightarrow 2 \times 6 - d = \frac{21}{2} \left( \text { From }(3) \right)\]
\[ \Rightarrow d = \frac{3}{2}\]
\[\text { Putting the value in (3), we get: } \]
\[n = 4\]
\[ \Rightarrow 2n = 8\]
\[\text { Thus, there are 8 terms in the progression } . \]
\[\text { To find the value of the first term: } \]
\[ a_2 + a_4 + . . . + a_{2n} = 30\]
\[ \Rightarrow (a + d) + (a + 3d) + . . . + [a + (2n - 1)d] = 30\]
\[ \Rightarrow \frac{n}{2}\left[ \left( a + d \right) + a + (2n - 1)d \right] = 30\]
\[\text { Putting n = 4 and d }= \frac{3}{2}, \text { we get: } \]
\[ a = \frac{3}{2}\]
\[\text { So, the series will be } 1\frac{1}{2}, 3, 4\frac{1}{2} . . .\]
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