From 75a5c861e984283f4c7b59f5ef3719d4103b0cc1 Mon Sep 17 00:00:00 2001
From: Zhao Xin <7176466@qq.com>
Date: Wed, 7 Sep 2022 21:40:33 +0800
Subject: [PATCH] gitpush

---
 问题/等差数列.md | 6 ++----
 1 file changed, 2 insertions(+), 4 deletions(-)

diff --git a/问题/等差数列.md b/问题/等差数列.md
index c4df8d5..710b57d 100644
--- a/问题/等差数列.md
+++ b/问题/等差数列.md
@@ -9,12 +9,10 @@
 
 $$ a_n = a + (n-1)d \tag{等差数列通项公式} $$
 
-换句话说，任意一个等差数列 $\{a_{n}\}$ 都可以写成：$\{\,a,\,a+d,\,a+2d,\,\cdots,\,a+(n-1)d\,\}$
-
 由等差数列通项公式可以得到以下公式：
 
 $$ n = \dfrac{a_n - a}{d} + 1 \tag{已知首项尾项及公差求项数}$$
 
-等差数列求和公式：
+换句话说，任意一个等差数列 $\{a_{n}\}$ 都可以写成：$\{\,a,\,a+d,\,a+2d,\,\cdots,\,a+(n-1)d\,\}$
 
-$$ \dfrac {(a + a_n) \times n}{2} $$
+$$ 等差数列各项的和 = \dfrac {(a + a_n) \times n}{2} $$