Suppose that the number of terms in an A.P. is $2 k, k \in N$. 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 :

Question

Aicharya · Accepted Answer

Let the A.P. be
$\begin{aligned}
& a, a+2, a+2 d, \ldots, a+(2 k-1) d \\ & \text { Now, } a+a+2 d+a+4 d+\ldots+a+(2 k-2) d=40 \\ & k a+2 d+4 d+\ldots+(2 k-2) d=40 \\ & \Rightarrow k a+\frac{k-1}{2}[2 d+2 k d-2 d]=40
\end{aligned}$
$\Rightarrow k a+k(k-1) d=40$ ...(1)
And $a+d+a+3 d+\ldots+a+(2 k-1) d=55$
$\Rightarrow k a+\frac{k}{2}(d+2 k d-d)=55$
$\Rightarrow k a+k^2 d=55$ ....(2)
Also, $a+(2 k-1) d-a=27$
$\Rightarrow \quad(2 k-1) d=27 \Rightarrow d=\frac{27}{2 k-1}$ ...(3)
From equation (1) and (2)
$k^2 d-k d-k^2 d=-15$
$\Rightarrow \quad d=\frac{15}{k}$
From equation (3) and (4)
$\begin{aligned}
& \frac{27}{2 k-1}=\frac{15}{k} \\ & 27 k=30 k-15 \\ & \Rightarrow 3 k=15 \\ & \Rightarrow k=5
\end{aligned}$ ^

Suppose that the number of terms in an A.P. is $2 k, k \in N$. If the sum of all odd terms of the A.P. is 40…

Solution