gaokao 2021 Q17
Multi-Part Structured Problem on AP
17. Let $S _ { n }$ denote the sum of the first $n$ terms of an arithmetic sequence $\left\{ a _ { n } \right\}$ with non-zero common difference. If $a _ { 3 } = S _ { 5 }$ and $a _ { 2 } a _ { 4 } = S _ { 4 }$ .
(1) Find the general term formula $a _ { n }$ of the sequence $\left\{ a _ { n } \right\}$ ;
(2) Find the minimum value of $n$ such that $S _ { n } > a _ { n }$ holds.
Answer: (1) $a _ { n } = 2 n - 6$ ; (2) 7 .
[Solution]
[Analysis] (1) From the given conditions, first find the value of $a _ { 3 }$ , then combine with the given conditions to find the common difference of the sequence to determine the general term formula;
(2) First find the expression for the sum of the first $n$ terms, then solve the quadratic inequality to determine the minimum value of $n$ . [Detailed Solution] (1) By the properties of arithmetic sequences, we have $S _ { 5 } = 5 a _ { 3 }$ , thus: $a _ { 3 } = 5 a _ { 3 }$ , therefore $a _ { 3 } = 0$ ,
Let the common difference of the arithmetic sequence be $d$ . Then: $a _ { 2 } a _ { 4 } = \left( a _ { 3 } - d \right) \left( a _ { 3 } + d \right) = - d ^ { 2 }$ , $S _ { 4 } = a _ { 1 } + a _ { 2 } + a _ { 3 } + a _ { 4 } = \left( a _ { 3 } - 2 d \right) + \left( a _ { 3 } - d \right) + a _ { 3 } + \left( a _ { 3 } + d \right) = 4a_3 - 2d = -2d$ ,
Thus: $- d ^ { 2 } = - 2 d$ . Since the common difference is non-zero, we have: $d = 2$ ,
The general term formula of the sequence is: $a _ { n } = a _ { 3 } + ( n - 3 ) d = 2 n - 6$ .
(2) From the general term formula, we have: $a _ { 1 } = 2 - 6 = - 4$ , thus: $S _ { n } = n \times ( - 4 ) + \frac { n ( n - 1 ) } { 2 } \times 2 = n ^ { 2 } - 5 n$ . The inequality $S _ { n } > a _ { n }$ becomes: $n ^ { 2 } - 5 n > 2 n - 6$ . Simplifying: $( n - 1 ) ( n - 6 ) > 0$ ,
Solving: $n < 1$ or $n > 6$ . Since $n$ is a positive integer, the minimum value of $n$ is 7 . [Key Point] The solution of basic quantities in arithmetic sequences is a fundamental problem in arithmetic sequences. The key to solving such problems is to master the relevant formulas of arithmetic sequences and apply them flexibly.