If the sum of the first n terms of an AP is $4n – n^2 $, what is the first term (that is $S_1$)? What is the sum of first two terms? What is the second term? Similarly, find the $3^{rd}$, the $10^{th}$ and the $n^{th}$ terms.

Question

For positive integers $ n $, if $ 4 a_n = \frac{n^2 + 5n + 6}{4} $ and $$ S_n = \sum_{k=1}^{n} \left( \frac{1}{a_k} \right), \text{ then the value of } 507 S_{2025} \text{ is:} $$

Answer 1

Given that $S_n = 4n − n^2$, the first term is $a_1 = S_1 = 4(1) − (1)^2 = 3$. The sum of the first two terms is $S_2 = 4(2) − (2)^2 = 4$. Therefore, the second term is $a_2 = S_2 − S_1 = 4 − 3 = 1$. The common difference is $d = a_2 − a_1 = 1 − 3 = −2$. The formula for the n^th term is $a_n = a + (n − 1)d$, which becomes $a_n = 3 + (n − 1) (−2)$. Simplifying this gives $a_n = 3 − 2n + 2$, so $a_n = 5 − 2n$. Consequently, the 3^rd term is $a_3 = 5 − 2(3) = −1$ and the 10^th term is $a_{10} = 5 − 2(10) = −15$.

The sum of the first two terms is 4. The second term is 1. The 3^rd, 10^th, and n^th terms are −1, −15, and 5−2n, respectively.

If the sum of the first n terms of an AP is \(4n – n^2 \), what is the first term (that is \(S_1\))? What is the sum of first two terms? What is the second term? Similarly, find the \(3^{rd}\), the \(10^{th}\) and the \(n^{th}\) terms.

Solution and Explanation

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