Suppose that the number of terms in an A.P.

Question

Suppose that the number of terms in an A.P. is $2k$, $k\in\mathbb{N}$. If the sum of all odd-position terms of the A.P. is $40$, the sum of all even-position terms is $55$, and the last term exceeds the first term by $27$, then $k$ is equal to:

Accepted Answer

With common difference $d$ (of the original A.P.), the odd-position and even-position terms each form an A.P. of $k$ terms with common difference $2d$:
$$\frac{k}{2}\big[2a_1+(k-1)2d\big] = 40, \qquad \frac{k}{2}\big[2a_2+(k-1)2d\big] = 55.$$

Also $a_{2k}-a_1 = 27 \Rightarrow d = \dfrac{27}{2k-1}$.

Subtracting the two sum equations gives $d = \dfrac{15}{k}$ (since the second exceeds the first by $k\cdot d = 15$). Equating: $\dfrac{27}{2k-1}=\dfrac{15}{k} \Rightarrow 27k = 15(2k-1) \Rightarrow 27k=30k-15 \Rightarrow 3k=15 \Rightarrow k=5.$

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