From 12d67021c96cb528956e8167f81a9d40a94f4a4e Mon Sep 17 00:00:00 2001
From: Zhao Xin <7176466@qq.com>
Date: Wed, 7 Sep 2022 21:33:25 +0800
Subject: [PATCH] gitpush

---
 问题/等差数列.md | 6 +++++-
 1 file changed, 5 insertions(+), 1 deletion(-)

diff --git a/问题/等差数列.md b/问题/等差数列.md
index 4df44b3..2091185 100644
--- a/问题/等差数列.md
+++ b/问题/等差数列.md
@@ -13,4 +13,8 @@ $$ a_n = a + (n-1)d \tag{等差数列通项公式} $$
 
 由等差数列通项公式可以得到以下公式：
 
-$$ n = \dfrac{a_末-a}{d} + 1 \tag{已知首项尾项及公差求项数}$$
+$$ n = \dfrac{a_n}{d} + 1 \tag{已知首项尾项及公差求项数}$$
+
+等差数列求和公式：
+
+$$ \dfrac {(a + a_n) \times n}{2} $$