In an AP:
given a = 5, d = 3, an = 50, find n and S_n
given a = 7, a₁₃ = 35, find d and S₁₃.
given a₁₂ = 37, d = 3, find a and S₁₂.
given a₃ = 15, S₁₀ = 125, find d and a₁₀.
given d = 5, S₉ = 75, find a and a₉.
given a = 2, d = 8, S_n = 90, find n and a_n .
given a = 8, a_n = 62, S_n = 210, find n and d.
given a_n = 4, d = 2, S_n = –14, find n and a.
given a = 3, n = 8, S = 192, find d.
given l = 28, S = 144, and there are total 9 terms. Find a.

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

(i) Given: $a = 5$, $d = 3$, $a_n = 50$. Using the formula $a_n = a + (n − 1)d$, we substitute the given values: $50 = 5 + (n − 1)3$. Simplifying, we get $45 = (n − 1)3$, then $15 = n − 1$, which leads to $n = 16$. The sum $S_n$ is calculated using $S_n = \frac{n}{2}[a + a_n]$. Substituting $n=16$, $a=5$, and $a_n=50$, we get $S_{16} = \frac {16}{2}[5 + 50]$. This simplifies to $S_{16} = 8 \times 55$, resulting in $S_{16} = 440$.

(ii) Given: $a = 7$, $a_{13} = 35$. Using the formula $a_n = a + (n − 1) d$, we have $a_{13} = a + (13 − 1) d$. Substituting the given values, $35 = 7 + 12 d$. Rearranging, $35 − 7 = 12d$, so $28 = 12d$. Therefore, $d = \frac {28}{12}$, which simplifies to $d = \frac 73$. The sum $S_n$ is calculated using $S_n = \frac n2[a + a_n]$. For $n=13$, we have $S_{13} =\frac n2[a + a_{13}]$. Substituting the values, $S_{13}= \frac {13}{2}[7 + 35]$. This becomes $S_{13} = \frac {13 \times 42}{2}$, which simplifies to $S_{13} = 13 \times 21$, resulting in $S_{13} = 273$.

(iii) Given: $a_{12} = 37$, $d = 3$. Using the formula $a_n = a + (n − 1)d$, we have $a_{12}= a + (12 − 1)3$. Substituting the given value, $37 = a + 33$. Therefore, $a = 4$. The sum $S_n$ is calculated using $S_n = \frac n2[a + a_n]$. For $n=12$, $S_{12} = \frac {12}{2}[4 + 37]$. This simplifies to $S_{12}= 6 \times 41$, resulting in $S_{12} = 246$.

(iv) Given: $a_3 = 15$, $S_{10} = 125$. Using the formula $a_n = a + (n − 1)d$, we get $a_3 = a + (3 − 1)$, so $15 = a + 2d$ (Equation i). Using the formula $S_n = \frac {n}{2}[2a + (n-1)d]$, we have $S_{10} = \frac {10}{2}[2a + (10-1)d]$. Substituting $S_{10}=125$, we get $125 = 5[2a + 9d]$, which simplifies to $25 = 2a + 9d$ (Equation ii). Multiply Equation (i) by 2: $30 = 2a + 4d$ (Equation iii). Subtract Equation (iii) from Equation (ii): $25 - 30 = (2a + 9d) - (2a + 4d)$, which gives $-5 = 5d$, so $d = −1$. Substitute $d = −1$ into Equation (i): $15 = a + 2(−1)$, so $15 = a − 2$, which means $a = 17$. To find $a_{10}$, we use $a_{10} = a + (10 − 1)d$. Substituting the values, $a_{10}= 17 + (9) (−1)$, so $a_{10} = 17 − 9 = 8$.

(v) Given: $d = 5$, $S_9 = 75$. Using the formula $S_n = \frac {n}{2}[2a + (n-1)d]$, we have $S_9 =\frac 92[2a + (9-1)5]$. Substituting $S_9=75$, we get $75 =\frac 92(2a + 40)$. Simplifying, $25 = 3(a + 20)$, so $25 = 3a + 60$. Rearranging, $3a = 25 − 60$, which gives $a = -\frac {35}{3}$. To find $a_9$, we use $a_n = a + (n − 1)d$. Substituting the values, $a_9 = -\frac {35}{3} + (9 − 1)5$. This becomes $a_9 = -\frac {35}{3} + 8 \times 5$, so $a_9 = -\frac {35}{3} + 40$. Combining these, $a_9 = \frac {-35+120}{3}$, resulting in $a_9 = \frac {85}{3}$.

(vi) Given: $a = 2$, $d = 8$, $S_n = 90$. Using the formula $S_n = \frac {n}{2}[2a + (n-1)d]$, we have $90 = \frac n2[2 \times 2 + (n-1)8]$. Simplifying, $90 = \frac n2[4 + (n-1)8]$. Further simplification yields $90 = n [2 + (n − 1)4]$, so $90 = n [2 + 4n − 4]$. This leads to $90 = n (4n − 2)$, or $90= 4n^2 − 2n$. Rearranging into a quadratic equation: $4n^2 − 2n − 90 = 0$. Factoring the equation, we get $4n^2 − 20n + 18n − 90 = 0$, which factors as $4n (n − 5) + 18 (n − 5) = 0$, and then $(n − 5) (4n + 18) = 0$. The possible solutions for $n$ are $n − 5 = 0$ or $4n + 18 = 0$. This gives $n = 5$ or $n = -\frac {18}{4} = -\frac {9}{2}$. Since $n$ must be a positive integer, we conclude $n = 5$. To find $a_n$, we use $a_n = a + (n − 1)d$. For $n=5$, $a_5 = 2 + (5 − 1)8$. This calculates to $a_5= 2 + 4 \times 8$, so $a_5= 2 + 32$, resulting in $a_5 = 34$.

(vii) Given: $a = 8$, $a_n = 62$, $S_n = 210$. Using the formula $S_n = \frac {n}{2}[a + a_n]$, we substitute the given values: $210 = \frac n2[8 + 62]$. Simplifying, $210 = \frac n2 \times 70$. To find $n$, we solve $210 = 35n$, so $n = 6$. Using the formula $a_n = a + (n − 1)d$, we have $62 = 8 + (6 − 1)d$. Rearranging, $62 − 8 = 5d$, so $54 = 5d$. Therefore, $d = \frac {54}{5}$.

(viii) Given: $a_n = 4$, $d = 2$, $S_n = −14$. Using the formula $a_n = a + (n − 1)d$, we get $4 = a + (n − 1)2$. Simplifying, $4 = a + 2n − 2$, so $a + 2n = 6$. This gives $a = 6 − 2n$ (Equation i). Using the formula $S_n = \frac n2[a + a_n]$, we have $-14 = \frac n2[a + 4]$. Multiplying by 2, $-28 = n (a + 4)$. Substitute the expression for $a$ from Equation (i): $-28 = n (6 − 2n + 4)$. Simplifying, $-28 = n (− 2n + 10)$, so $-28 = − 2n^2 + 10n$. Rearranging into a quadratic equation: $2n^2 − 10n − 28 = 0$. Dividing by 2, $n^2 − 5n −14 = 0$. Factoring the equation: $n^2 − 7n + 2n − 14 = 0$, which factors as $n (n − 7) + 2(n − 7) = 0$, and then $(n − 7) (n + 2) = 0$. The possible solutions for $n$ are $n − 7 = 0$ or $n + 2 = 0$. This gives $n = 7$ or $n = −2$. Since $n$ must be a positive integer, we conclude $n = 7$. Substitute $n = 7$ into Equation (i) to find $a$: $a = 6 − 2 \times 7$. This calculates to $a= 6 − 14$, resulting in $a= −8$.

(ix) Given: $a = 3$, $n = 8$, $S = 192$. Using the formula $S_n = \frac n2[2a + (n-1)d]$, we substitute the given values: $192 = \frac 82[2 \times 3 + (8-1)d]$. Simplifying, $192 = 4 [6 + 7d]$. Dividing by 4, $48 = 6 + 7d$. Rearranging, $42 = 7d$, so $d = 6$.

(x) Given: $l = 28$, $S = 144$, and there are $9$ terms. Using the formula $S_n =\frac n2(a+l)$, we substitute the given values: $144 = \frac 92(a+28)$. Multiplying by 2 and dividing by 9, $16 = \frac 12(a+28)$. Multiplying by 2, $32 = a + 28$. Rearranging to solve for $a$, $a = 32 - 28$, so $a = 4$.

Solution and Explanation

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