# Given that a6 + a9 = 90 - a1 determine a1 + a2 +..+ a10 if a1, a2, ... are the terms of an a.p.Given that a6 + a9 = 90 - a1 determine a1 + a2 +..+ a10 if a1, a2, ... are the terms of an a.p.

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The general term of an AP is a+ (n-1)d where a is the first term and d is the common difference.

We have a6 + a9 = 90 - a1

=> a + 5d + a + 8d = 90 - a

=> 3a + 13d =90

=> a = (90-13d) / 3

Now the sum of the first n terms of an AP is given by a+ n*(n-1)*d/2

So the sum from a1 to a10 is a+ 9*10*d/2

= a + 45 d

=> (90-13d) / 3 +45d

=> 90/3 - 13d/3 +45d

=> 30 + (122/3)*d

We know that the rth term of the AP is given by ar = a1+(r-1).

Given a6+a9 = 90-a1.

So a1+a6+a9 = 90

a1 +a1+(5-1)d +a1+(9-1)d = 90

3a1+9d = 90.

3a1 = 90-9d, Or 9d = (90-3a1. Or d =(90-3a1)/9 = (30 -a1)/3.

a1 = (90-9d)/3 = 3(10-d) = 30-3d.

a10 = 30-3d + 9d. 30-6d.

Therefore , a1+a2+...+a10 = (1st term+last term)*number of terms)/2 = (a1+a1+9d)10/2 = {2a1+9 (30-a1)/3}10/5

= {2a1 +(90-3a1)}5

= 450 - 5a1.

If a1, a2 ..., a10 are the terms of an arithmetical progression, then their sum is:

S = (a1+a10)*10/2

We have, from enunciation, the constraint:

a3+a6 = 90-a9

We also know that each term of an arithmetical progression could be obtained from the previous term, by adding a constant amount.

So:

a3 = a2+d = a1+d+d=a1+2d

a6 = a1 + 5d

a9 = a1 + 8d

Now, we'll substitute them into the given relation:

a3+a6 = 90-a9

a1 + 2d + a1 + 5d = 90 - (a1 + 8d)

We'll move all terms in a1 and d, to the left side and we'll combine like terms:

3a1 + 15d = 90

We'll factorize by 3:

3(a1 + 5d) = 90

We'll divide by 3:

a1 + 5d = 30

Now, we'll consider the sum:

S = (a1+a10)*10/2

a10 = a1 + 9d

a1+a10 = a1 + a1 + 9d = 2a1 + 9d = a1+5d+a1+5d-d

But a1 + 5d = 30, so:

a1+a10 = 2*30 - d

S = (a1+a10)*10/2

S = (60-d)*10/2

S = 5(60-d)

S = 300 - 5d