Suppose that the number of terms in an A.P.\ is $2k$, $k\in\mathbb{N}$. If the sum of all odd terms of the A.P.\ is $40$, the sum of all even terms is $55$, and the last term of the A.P.\ exceeds the first term by $27$, then $k$ is equal to:
Step-by-Step Solution
Key Concept: Sum of even-positioned terms minus sum of odd-positioned terms in a $2k$-term A.P.\ equals $kd$. Combined with $(2k-1)d=27$, this is two equations in $k,d$.
Terms: $a, a+d, a+2d,\dots,a+(2k-1)d.$ Odd-positioned ($k$ terms): $a, a+2d,\dots, a+(2k-2)d.$
Sum of odd-positioned $=ka+k(k-1)d=40.$ Sum of even-positioned $=ka+k^{2}d=55.$
Subtracting gives $kd=15$, i.e.\ $d=15/k.$
$(2k-1)d=27\Rightarrow (2k-1)\dfrac{15}{k}=27\Rightarrow 30k-15=27k\Rightarrow k=5.$
Correct Answer: 2