Let a1,a2,a3,… be in an A.P. such that k=1∑12a2k−1=−572a1, a1=0. If k=1∑nak=0, then n is:
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Step-by-Step Solution
Step 1: Express the sum of odd terms
The given sum involves terms with odd indices: a1,a3,…,a23. This is an arithmetic progression with 12 terms, where the first term is a1 and the common difference is 2d (since we are skipping terms).
Step 2: Calculate the sum of the odd terms
Using the formula for the sum of an arithmetic progression, Sm=2m(2A+(m−1)D), where m=12, A=a1, and D=2d. We substitute these values to find the sum of the odd terms.
Step 3: Formulate an equation for the common difference
We equate the calculated sum of odd terms to the given expression, −572a1. This allows us to establish a relationship between the first term a1 and the common difference d. We simplify the equation to find a1 in terms of d.
Step 4: Set up the sum of the first n terms
The problem states that the sum of the first n terms of the arithmetic progression is 0. We use the formula for the sum of an arithmetic progression, Sn=2n(2a1+(n−1)d), and set it equal to 0.
Step 5: Solve for n
Substitute a1=−5d into the equation for the sum of the first n terms. Since a1=0, it implies d=0. Also, n cannot be 0 for a sum of terms. Therefore, the only possibility for the sum to be 0 is if the term (n−11) is 0.
Step 6: Determine the value of n
From the previous step, we deduced that n−11=0. Solving for n gives us the final answer.