Let 
, 
, . . . , 
be an arithmetic sequence. If 
and 
, then find .
1. arithmetic sequence: $$\small{\text{
$
a_1, a_2,a_3,a_4,a_5,a_6,a_7,a_8,a_9,a_{10}
$
}}$\\$
\small{\text{
$
a_n=a_1+(n-1)*d $ and
$
s_n= \frac{n}{2}*[2a_1*(n-1)d]
$
}}
\small{\text{
$
\Rightarrow s_{10}= 17+15 = \frac{10}{2}*[2a_1+(10-1)d]
$
}} $\\$
\small{\text{
$
32 = 5*(2a_1+9d)
$
}} $\\$
\small{\text{
$
\textcolor[rgb]{0,0,1}{(1) \quad 32 = 10a_1+45d}
$
}}$$
2. arithmetic sequence:
$$\small{\text{
$
a_1,a_3,a_5,a_7,a_9$
}}$\\$
\small{\text{
$
a_i=a_1+(i-1)*(2d) $ and
$
s_i= \frac{i}{2}*[2a_1*(i-1)(2d)]
$
}}
\small{\text{
$
\Rightarrow s_5= 17 = \frac{5}{2}*[2a_1+(5-1)(2d)]
$
}} $\\$
\small{\text{
$
2*17 = 5*(2a_1+8d)
$
}} $\\$
\small{\text{
$
\textcolor[rgb]{0,0,1}{(2) \quad 34 = 10a_1+40d}
$
}}$$
$$\small{\text{
(1) - (2):
$\quad
32-34 = 10a_1-10a_1 + 45d-40d = 5d \qquad -2=5d \qquad \boxed{d = -\frac{2}{5}}
$
}}$$
$$\small{\text{
d into (2):
$
\quad
34 = 10a_1+40*(-\frac{2}{5}) = 10a_1 - 16 \qquad 10a_1 = 50 \qquad \boxed{a_1 =5 }
$
}}$$
.
