Question 936448
Hello,
We are working on Arithmetic Sequences. I am stuck.
Can someone please help. I need to find a1 and d.

a15 = 129, a8=22(a1)

Thanks.
<pre><font face = "Tohoma" size = 4 color = "indigo"><b>If what you're saying is that: {{{a[15] = 129}}}, and {{{a[8] = 22(a[1])}}}, then:
1st term, or {{{highlight_green(a[1] = 3)}}}, and common difference, or {{{highlight_green(d = 9)}}}

Since you confirmed that it is in fact: {{{a[15] = 129}}}, and {{{a[8] = 22(a[1])}}}, then the solution follows:
{{{a[n] = a[1] + (n - 1)d }}}
{{{a[15] = a[1] + (15 - 1)d}}}
{{{a[15] = a[1] + 14d}}}
{{{129 = a[1] + 14d}}} ------ Substituting 129 for {{{a[15]}}} 
{{{a[1] = 129 - 14d}}} ------- eq (i)

{{{a[n] = a[1] + (n - 1)d}}} 
{{{a[8] = a[1] + (8 - 1)d}}}
{{{a[8] = a[1] + 7d}}}
{{{22(a[1]) = a[1] + 7d}}} ------ Substituting {{{22(a[1])}}} for {{{a[8]}}} 
{{{22a[1] - a[1] = 7d}}} 
{{{21a[1] = 7d}}} ------- eq (ii)
21(129 – 14d) = 7d --------- Substituting 129 – 14d for {{{a[1]}}} in eq (ii)
2,709 – 294d = 7d
7d + 294d = 2,709
301d = 2,709
d, or common difference = {{{2709/301}}}, or {{{highlight_green(9)}}}

{{{21a[1] = 7(9)}}} -------- Substituting 9 for d in eq (ii)
{{{21a[1] = 63}}}
1st term, or {{{a[1] = 63/21}}}, or {{{highlight_green(3)}}}
You can do the check!! 
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