解题思路::
a
2
+
a
5
+
a
17
+
a
22
b
8
+
b
10
+
b
12
+
b
16
=
(
a
2
+
a
22
) +(
a
5
+
a
17
)
(
b
8
+
b
16
) +(
b
10
+
b
12
)
=
2
a
12
+2
a
11
2
b
12
+2
b
11
=
S
22
T
22
.由此能求出其具体结果.
a2+a5+a17+a22
b8+b10+b12+b16=
(a2+a22) +(a5+a17)
(b8+b16) +(b10+b12)
=
2a12+2a11
2b12+2b11=
a1+a22
b1+b22
=
22
2(a1+a22)
22
2(b1+b22) =
S22
T22=[7×22+1/22+3]=[31/5].
故答案:[31/5].
点评:
本题考点: 等差数列的性质.
考点点评: 本题考查等差数列的性质和应用,解题时要注意公式的合理应用.