Suppose that the number of terms in an A.P. is 2k,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:
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Step-by-Step Solution
Step 1: Define the A.P. and its terms
Let the arithmetic progression (A.P.) be denoted by a1,a2,…,a2k, where a1 is the first term and d is the common difference. The total number of terms is 2k.
Step 2: Formulate equations from given sums
The odd terms form an A.P. with k terms, first term a1, and common difference 2d. The sum of odd terms is given as 40. The even terms also form an A.P. with k terms, first term a2=a1+d, and common difference 2d. The sum of even terms is given as 55. We use the sum formula Sn=2n(2a+(n−1)d′).
Step 3: Subtract the sum equations
Subtracting equation (1) from equation (2) simplifies the expression significantly, allowing us to find a relationship between k and d.
Step 4: Solve for kd
After subtracting the equations, most terms cancel out, leaving us with a simple equation kd=15. This is a crucial relationship.
Step 5: Use the last term condition
The problem states that the last term, a2k, exceeds the first term, a1, by 27. We use the formula for the n -th term of an A.P., an=a1+(n−1)d, to express a2k.
Step 6: Solve for k
Substituting kd=15 from equation (3) into the simplified last term condition, we get 2(15)−d=27. This allows us to solve for d and then for k.
Step 7: Calculate d and k
From 30−d=27, we find d=3. Then, using kd=15, we substitute d=3 to find k=5.