Question

Let $a_1$, $a_2$, . . . , $a_{10}$ be an arithmetic sequence. if $a_1 + a_3 + a_5 + a_7 + a_9 = 17$ and $a_2 + a_4 + a_6 + a_8 + a_{10} = 15$, what is the value of $a_1$?
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Answer 1

Because $a_1$ through $a_{10}$ are in arithmetic progression, there is some fixed number $d$ between consecutive terms. This means we have

$a_2 = a_1 + d$

$a_3 = a_2 + d = a_1 + 2d$

$a_4 = a_3 + d = a_1 + 3d$

and so on, with the general $n$-th term

$a_n = a_1 + (n-1) d$

Then we can write each given equation in terms of $a_1$ and $d$ :

$a_1 + a_3 + a_5 + a_7 + a_9 = 17$

$\implies a_1 + (a_1+2d) + (a_1+4d) + (a_1 + 6d) + (a_1 + 8d) = 17$

$\implies 5a_1 + 20d = 17$

and

$a_2 + a_4 + a_6 + a_8 + a_{10} = 15$

$\implies (a_1+d) + (a_1+3d) + (a_1+5d) + (a_1+7d) + (a_1+9d) = 15$

$\implies 5a_1+25d= 15$

Eliminate $d$ and solve for $a_1$ :

$5 (5a_1 + 20d) - 4 (5a_1 + 25d) = 5\times17-4\times15$

$\implies 5a_1 = 25$

$\implies \boxed{a_1 = 5}$

Answer 2

Answer: a1=5

Step-by-step explanation: