Suppose that the number of terms in an A.P. is $2k$, $k \in N$. 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:
- (1) 5
- (2) 8
- (3) 6
- (4) 4
Solution:
Let the A.P. be $a_1, a_2, \dots, a_{2k}$. Common difference is $d$.
Sum of odd terms ($k$ terms):
$$ S_{odd} = a_1 + a_3 + \dots + a_{2k-1} = 40 $$
Sum of even terms ($k$ terms):
$$ S_{even} = a_2 + a_4 + \dots + a_{2k} = 55 $$
Subtracting the two sums:
$$ S_{even} – S_{odd} = (a_2 – a_1) + (a_4 – a_3) + \dots + (a_{2k} – a_{2k-1}) $$
$$ 55 – 40 = d + d + \dots + d \quad (k \text{ times}) $$
$$ 15 = kd \Rightarrow d = \frac{15}{k} $$
Given last term exceeds first by 27:
$$ a_{2k} – a_1 = 27 $$
$$ a_1 + (2k-1)d – a_1 = 27 $$
$$ (2k-1)d = 27 $$
Substituting $d = \frac{15}{k}$:
$$ (2k-1)\frac{15}{k} = 27 $$
$$ 15(2k-1) = 27k $$
$$ 30k – 15 = 27k $$
$$ 3k = 15 \Rightarrow k = 5 $$
Ans. (1)