Let Sn=21+61+121+201+… up to n terms. If the sum of the first six terms of an A.P. with first term −p and common difference p is 2026S2025, then the absolute difference between 20th and 15th terms of the A.P. is
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Step-by-Step Solution
Step 1: Find the general term of the series Sn
First, we need to find the general term of the given series Sn. We can observe that the denominators are products of consecutive integers: 1×2, 2×3, 3×4, 4×5, and so on. So, the kth term, Tk, can be written as k(k+1)1. This term can be decomposed using partial fractions into k1−k+11.
Step 2: Calculate the sum Sn
Now, we calculate the sum Sn by summing the general terms. This is a telescoping series, where most terms cancel out. The sum simplifies to 1−n+11, which can be written as n+1n.
Step 3: Calculate S2025
Using the formula for Sn we just derived, we can find S2025 by substituting n=2025. This gives us S2025=20262025.
Step 4: Find the sum of the first six terms of the A.P.
For the A.P., the first term is a=−p and the common difference is d=p. The sum of the first six terms, S6, is given by the formula Sn=2n(2a+(n−1)d). Substituting the values, we get S6=9p. We are given that S6=2026S2025. Substituting the value of S2025, we find that 9p=45, which means p=5.
Step 5: Calculate the absolute difference between the 20th and 15th terms
We need to find the absolute difference between the 20th and 15th terms of the A.P. The nth term of an A.P. is an=a+(n−1)d. So, a20=a+19d and a15=a+14d. The difference is a20−a15=5d. Since d=p and we found p=5, the difference is 5×5=25. The absolute difference is ∣25∣=25.