Let $a_1,$ $a_2,$ $a_3,$ $\dots,$ $a_8,$ $a_9,$ $a_{10}$ be an arithmetic sequence. If $a_1 + a_3 + a_5 + a_7 + a_9 = 5$ and $a_2 + a_4 + a_6 + a_8 + a_{10} = -2$, then find $a_1$.
a1 + a3 + a5 + a7 + a9 = +5 (call this equation 1)
a2 + a4 + a6 + a8 + a10 = – 2 (call this equation 2)
subtract (eq 1) from (eq 2) to obtain
a2 – a1 + a4 – a3 + a6 – a5 + a8 – a7 + a10 – a9 = – 7
add parentheses for clarity to show relationship
(a2 – a1) + (a4 – a3) + (a6 – a5) + (a8 – a7) + (a10 – a9) = – 7
Let the common difference between terms be called "d"
remember that (a2 – a1) = d and so on for (a4 – a3) and the rest
(a2 – a1) + (a4 – a3) + (a6 – a5) + (a8 – a7) + (a10 – a9) = – 7
d + d + d + d + d = – 7
5d = – 7
d = – 7 / 5
per (eq 1) a1 + (a1 + 2d) + (a1 + 4d) + (a1 + 6d) + (a1 + 8d) = 5
a1 + (a1 – 14 / 5) + (a1 – 28 / 5) + (a1 – 42 / 5) + (a1 – 56 / 5) = 5
5a1 + (– 140 / 5) = 5
5a1 – 28 = 5
5a1 = 33
a1 = 33 / 5
check answer
from (eq 2) a2 + a4 + a6 + a8 + a10 = – 2 and we know that a2 = a1 – 7 / 5 = 26 / 5
(26 / 5) + (26 / 5 – 14 / 5) + (26 / 5 – 28 / 5) + (26 / 5 – 42 / 5) + (26 / 5 – 56 / 5) =? – 2
(5)(26 / 5) + (– 140 / 5) =? – 2
(130 / 5) – (140 / 5) =? – 2
– 10 / 5 =? – 2
– 2 = – 2 true
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