Suppose that the number of terms in an A.P. is 2 k, k 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

Let the A.P. be aligned & a, a+2, a+2 d, , a+(2 k-1) d & Now, a+a+2 d+a+4 d+ +a+(2 k-2) d=40 & k a+2 d+4 d+ +(2 k-2) d=40 & k a+ k-1 2 [2 d+2 k d-2 d]=40 aligned k a+k(k-1) d=40 ...(1) And a+d+a+3 d+ +a+(2 k-1) d=55 k a+ k 2 (d+2 k d-d)=55 k a+k^2 d=55 ....(2) Also, a+(2 k-1) d-a=27 (2 k-1) d=27 d= 27 2 k-1 ...(3) From equation (1) and (2) k^2 d-k d-k^2 d=-15 d= 15 k From equation (3) and (4) aligned & 27 2 k-1 = 15 k & 27 k=30 k-15 & 3 k=15 & k=5 aligned