Question

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

Accepted Answer

a6+a9 = 90 = a1
we know that:
a6 = a2+4r
==> a2+4r + a9 = 90-a1
==> a1+a2 + a9 + 4r = 90
==> a1+ a2 + a9 = 90-4r .......(1)
Now we need the values of:
a1+a2 + a10 = ??
a10 = a9 + r
==> a1 + a2 + a10 = a1 + a2 + a9 + r
              From (1) :
                               = 90-4r + r
                                = 90-3r

Answer

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

Answer

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.

Answer

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

Math

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

Expert Answers

We’ll help your grades soar

Related Questions

Determine the sum of the first 20 terms of an A.P. if a4 - a2 = 4 and a1 + a3 + a5 + a6 = 30.

Given that a5+a6=8 and r=2 find a1 if a1,a5,a6 are the terms of a.p.

Calculate a10 if a4 = 19, a1 = 10, a1 and a4 being the terms of A. P.

Calculate a1+...+a100, a1...a100 are terms of A.P. and a1=2, a2=4.

Write the first six terms of the sequence given by : a1=a2=1 an=2an-1 + 3an-2, n>2 n n.n-1,n-2 is an index.

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

Expert Answers

We’ll help your grades soar

Related Questions

Determine the sum of the first 20 terms of an A.P. if a4 - a2 = 4 and a1 + a3 + a5 + a6 = 30.

Given that a5+a6=8 and r=2 find a1 if a1,a5,a6 are the terms of a.p.

Calculate a10 if a4 = 19, a1 = 10, a1 and a4 being the terms of A. P.

Calculate a1+...+a100, a1...a100 are terms of A.P. and a1=2, a2=4.

Write the first six terms of the sequence given by : a1=a2=1 an=2*an-1 + 3*an-2, n>2 n n.n-1,n-2 is an index.

Write the first six terms of the sequence given by : a1=a2=1 an=2an-1 + 3an-2, n>2 n n.n-1,n-2 is an index.