The number of terms of an A.P.

Question

The number of terms of an A.P. is even; the sum of all the odd-position terms is $24$, the sum of all the even-position terms is $30$, and the last term exceeds the first by $\dfrac{21}{2}$. Then the number of terms of the A.P. which are integers is:

Accepted Answer

Let there be $2n$ terms with common difference $d$. The even-position terms exceed the corresponding odd-position terms by $nd$ in total: $30-24 = nd = 6$.

Last term $-$ first term: $(2n-1)d = \dfrac{21}{2} \implies 2nd - d = \dfrac{21}{2} \implies 12-d = \dfrac{21}{2} \implies d = \dfrac32.$ Then $n = 6/d = 4$, so there are $8$ terms total.

Sum of odd-position terms: $4a_1+12d = 24 \Rightarrow a_1 = 1.5$. The terms are $1.5, 3, 4.5, 6, 7.5, 9, 10.5, 12$ --- the terms at even positions ($3,6,9,12$) are integers: 4 terms.

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