Question 478502
Interior angles of an octagon ABCDEFGH are in A.P.the largest and second an average of 153.find the average of least two.
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The nth term of an arithmetic sequence (numbers in A.P.) is
given by the formula:

{{{a[n]=a[1]+(n-1)d}}}

The largest interior angle is the octagon is {{{a[8]}}}

{{{a[8]=a[1]+(8-1)d}}}

{{{a[8]=a[1]+7d}}}

The second interior angle is the octagon is {{{a[2]}}}

{{{a[2]=a[1]+(2-1)d}}}

{{{a[2]=a[1]+d}}}

The average of the largest and second interior angles is

{{{(a[1]+a[8])/2 = ((a[1]+7d)+(a[1]+d))/2=(2a[1]+8d)/2=(2(a[1]+4d))/2=a[1]+4d}}}

We are told this average is 153°

(eq. 1)  {{{a[1]+4d=153}}}

The sum of n terms of an arithmetic sequence (numbers in A.P.) is
given by the formula:

{{{S[n]=expr(n/2)(2a[1]+(n-1)d)}}}

Substituting n=8

{{{S[8]=expr(8/2)(2a[1]+(8-1)d)}}}

(eq. 2)   {{{S[8]=4(2a[1]+7d)}}}

The sum of the measurements of the interior angles of an n-sided
polygon is given by the epression 

{{{(n-2)*"180°"}}}

For an octagon, n=8.

Sum of the interior angles of the octagon is 

{{{(8-2)*"180°"=6*"180°"="1080°"}}}

Therefore {{{S[8]=1080}}}.  Substituting that in (eq. 2)

{{{1080=4(2a[1]+7d)}}}

Dividing both sides by 4

{{{270=2a[1]+7d}}} 

or

{eq. 3)   {{{2a[1]+7d=270}}}

So we have this system consisting of eqs. 1 and 3

{{{system(a[1]+4d=153, 2a[1]+7d=270)}}} 
 
Solve the first equation of the system for {{{a[1]}}}

(eq. 4)   {{{a[1]=153-4d}}}

Substituting in the second equation of the system

{{{2(153-4d)+7d=270}}}

{{{306-8d+7d=270}}}

{{{306-d=270}}}

{{{-d=-36}}}

{{{d=36}}}

Substituting in (eq. 4)

{{{a[1]=153-4(36)}}}

{{{a[1]=9}}}

So the least interior angle of the octagon is the first
term 9°, and the common difference is d=36°

So the second term {{{a[2]}}} = 9° + 36° = 45°.

The average of the least two interior angles is:

{{{(a[1]+a[2])/2=("9°"+"45°")/2="54°"/2="27°"}}}

Answer: 27°

Edwin<pre>