Question

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. 

Answer 1

Let total number of terms be 2n.
According to question, we have:

\[a_1 + a_3 + . . . + a_{2n - 1} = 24 . . . (1)\]

\[ a_2 + a_4 + . . . + a_{2n} = 30 . . . (2)\]

\[\text { Subtracting (1) from (2), we get: } \]

\[\left( d + d + . . . + \text { upto n terms } \right) = 6\]

\[ \Rightarrow nd = 6 . . . (3)\]

\[\text { Given }: \]

\[ a_{2n} = a_1 + \frac{21}{2}\]

\[ \Rightarrow a_{2n} - a_1 = \frac{21}{2}\]

\[ \Rightarrow a + (2n - 1)d - a = \frac{21}{2} [ \because a_{2n} = a + (2n - 1)d, a_1 = a]\]

\[ \Rightarrow 2nd - d = \frac{21}{2}\]

\[ \Rightarrow 2 \times 6 - d = \frac{21}{2} \left( \text { From }(3) \right)\]

\[ \Rightarrow d = \frac{3}{2}\]

\[\text { Putting the value in (3), we get: } \]

\[n = 4\]

\[ \Rightarrow 2n = 8\]

\[\text { Thus, there are 8 terms in the progression } . \]

\[\text { To find the value of the first term: } \]

\[ a_2 + a_4 + . . . + a_{2n} = 30\]

\[ \Rightarrow (a + d) + (a + 3d) + . . . + [a + (2n - 1)d] = 30\]

\[ \Rightarrow \frac{n}{2}\left[ \left( a + d \right) + a + (2n - 1)d \right] = 30\]

\[\text { Putting n = 4 and d  }= \frac{3}{2}, \text { we get: } \]

\[ a = \frac{3}{2}\]

\[\text { So, the series will be } 1\frac{1}{2}, 3, 4\frac{1}{2} . . .\]