Let , , . . . , be an arithmetic sequence. If and , then find .

Let , , . . . ,  be an arithmetic sequence. If  and , then find .

Subtracting the second sum from the first, we have

(a2 - a1) + (a4 - a3) + (a6 - a5) + (a8 - a7) + (a10 - a9) = -2   or

  d           +      d       +      d       +     d        +      d        =  -2   or

5d = -2     →   d = -2/5

Also, the sum of the arithmetic series is given by

S = (10/2)(a₁ + a₁₀) = 5(a₁ + a₁₀) = 32     divide both sides by 5

So

a₁ + a₁₀ = 32/5 →  a₁₀ = 32/5 - a₁  ( 1)

And the last term is given by

a₁₀ = a₁ + d(9)    (2)

Therefore, setting (1) = (2), we have

32/5 - a₁ = a₁ + d(9)     ... solve for a₁

(32/5 -2a₁) = d(9)

32/5 -2a₁ = (-2/5)(9) = -18/5  

32/5 + 18/5 = 2a₁

2a₁ = 50/5

a₁ = 5

And a₁₀=  32/5 - a₁ =  32/5 - 5 = 1.4

Proof

a₁ + a₃ + a₅ + a₇ + a₉  = 5 + 4.2 + 3.4 + 2.6 + 1.8 = 17

And since each of the other 5 terms  is (2/5) less than it's preceding term.....

[17 - 5(2/5)]  = [17 - 2] = 15  = the sum of the terms with even subscripts

Original sequence   a=a1    d=d

$$S_n=\frac{n}{2}[2a+(n-1)d]\qquad $This is the sum of an AP$$$

Solve simultaneously

(2)-(1)

d=3-17/5 = -2/5

sub into (2)

a₁+5(-2/5)=3

a₁-2=3

a₁=5

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:

$$\small{\text{
$
a_1,a_3,a_5,a_7,a_9$
}}$\\$
\small{\text{
$
a_i=a_1+(i-1)*(2d) $ and
$
s_i= \frac{i}{2}*[2a_1*(i-1)(2d)]
$
}}
\small{\text{
$
\Rightarrow s_5= 17 = \frac{5}{2}*[2a_1+(5-1)(2d)]
$
}} $\\$
\small{\text{
$
2*17 = 5*(2a_1+8d)
$
}} $\\$
\small{\text{
$
\textcolor[rgb]{0,0,1}{(2) \quad 34 = 10a_1+40d}
$
}}$$

 $$\small{\text{
(1) - (2):
$\quad
32-34 = 10a_1-10a_1 + 45d-40d = 5d \qquad -2=5d \qquad \boxed{d = -\frac{2}{5}}
$
}}$$

$$\small{\text{
d into (2):
$
\quad
34 = 10a_1+40*(-\frac{2}{5}) = 10a_1 - 16 \qquad 10a_1 = 50 \qquad \boxed{a_1 =5 }
$
}}$$