Let $a_1$, $a_2$, . . . , $a_{10}$ be an arithmetic sequence. If $a_1 + a_3 + a_5 + a_7 + a_9 = 17$ and $a_2 + a_4 + a_6 + a_8 + a_{10} = 15$, then find $a_1$.
+23246 Let the arithmetic sequence have first term a1 and common difference d.
Then: a2 = a1 + d
and: a3 = a1 + 2d
a4 = a1 + 3d
a5 = a1 + 4d
a6 = a1 + 5d
a7 = a1 + 6d
a8 = a1 + 7d
a9 = a1 + 8d
a10 = a1 + 9d
Since a1 + a3 + a5 + a7 + a9 = 17, then (a1) + (a1 + 2d) + (a1 + 4d) + (a1 + 6d) + (a1 + 8d) = 17.
Simplifying: 5a1 + 20d = 17
Since a2 + a4 + a6 + a8 + a10 = 15, then (a1 + d) + (a1 + 3d) + (a1 + 5d) + (a1 + 7d) + (a1 + 9d) = 15
Simplifying: 5a1 + 25d + 15
Combining: 5a1 + 20d = 17
5a1 + 25d = 15
Subtracting: -5d = 2
Dividing: d = -2/5
Substituting into 5a1 + 20d = 17 ---> 5a1 + 20(-2/5) = 17
---> 5a1 - 8 = 17
---> 5a1 = 25
---> a1 = 5
+23246 Let the arithmetic sequence have first term a1 and common difference d.
Then: a2 = a1 + d
and: a3 = a1 + 2d
a4 = a1 + 3d
a5 = a1 + 4d
a6 = a1 + 5d
a7 = a1 + 6d
a8 = a1 + 7d
a9 = a1 + 8d
a10 = a1 + 9d
Since a1 + a3 + a5 + a7 + a9 = 17, then (a1) + (a1 + 2d) + (a1 + 4d) + (a1 + 6d) + (a1 + 8d) = 17.
Simplifying: 5a1 + 20d = 17
Since a2 + a4 + a6 + a8 + a10 = 15, then (a1 + d) + (a1 + 3d) + (a1 + 5d) + (a1 + 7d) + (a1 + 9d) = 15
Simplifying: 5a1 + 25d + 15
Combining: 5a1 + 20d = 17
5a1 + 25d = 15
Subtracting: -5d = 2
Dividing: d = -2/5
Substituting into 5a1 + 20d = 17 ---> 5a1 + 20(-2/5) = 17
---> 5a1 - 8 = 17
---> 5a1 = 25
---> a1 = 5
