The number of terms of an A.P. is even; the sum of all the odd...

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 :

Options

A $4$
B $10$
C $6$
D $8$

Correct answer

A. $4$

Step-by-step solution

\mathrm{a}_2+\mathrm{a}_4+\ldots+\mathrm{a}_{\mathrm{n}}=30

...(1)<br/>

\mathrm{a}_1+\mathrm{a}_3+\ldots+\mathrm{a}_{\mathrm{n}-1}=24

...(2)<br/>

(1)-(2)

<br/>\left(a_2-a_1\right)+\left(a_4-a_3\right) \ldots\left(a_n-a_{n-1}\right)=6

\begin{aligned} & \Rightarrow \frac{\mathrm{n}}{2} \mathrm{~d}=6 \Rightarrow \mathrm{nd}=12 \\ & \mathrm{a}_{\mathrm{n}}-\mathrm{a}_1=(\mathrm{n}-1) \mathrm{d}=\frac{21}{2} \\ & \Rightarrow \mathrm{nd}-\mathrm{d}=\frac{21}{2} \Rightarrow 12-\frac{21}{2}=\mathrm{d} \\ & \Rightarrow \mathrm{d}=\frac{3}{2}, \mathrm{n}=8\end{aligned}

\begin{aligned}<br/>& \text{ Sum\text{ of } odd terms }=\frac{4}{2}[2 \mathrm{a}+(4-1) 3]=24 \\ & \Rightarrow \mathrm{a}=\frac{3}{2} \\ & \text{ A.P. } \Rightarrow \frac{3}{2}, 3, \frac{9}{2}, 6, \frac{15}{2}, 9, \frac{21}{2}, 12<br/>\end{aligned}

no. of integer terms

=4

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