Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

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 $a_2 = 14$ and $a_3 = 18$, the common difference $d$ is calculated as $d = a_3 - a_2 = 18 - 14 = 4$. Using the formula $a_2 = a + d$, we substitute the known values: $14 = a + 4$, which simplifies to $a = 10$. The sum of the first $n$ terms of an arithmetic progression is given by $Sn = \frac n2[2a + (n-1)d]$.

To find the sum of the first 51 terms, $S_{51}$, we use the formula: $S_{51} = \frac {51}{2} [2 \times 10 + (50-1)4]$.

This simplifies to $S_{51}= \frac {51}{2} [20 + (50)4]$.

Further calculation yields $S_{51}= \frac {51 \times 220}{2}$, which further reduces to $S_{51}= 51 \times 110$, resulting in $S_{51} = 5610$.

Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Solution and Explanation

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