SOLUTION: Let a_1, a_2, a_3, ..., a_8, a_9, a_{10} be an arithmetic sequence. If a_1 + a_3 + a_5 = 5 and a_2 + a_4 = -2, then find a_1.

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Question 1209375: Let a_1, a_2, a_3, ..., a_8, a_9, a_{10} be an arithmetic sequence. If a_1 + a_3 + a_5 = 5 and a_2 + a_4 = -2, then find a_1.
Found 3 solutions by ikleyn, math_tutor2020, greenestamps:
Answer by ikleyn(52790)   (Show Source): You can put this solution on YOUR website!
.
Let a_1, a_2, a_3, ..., a_8, a_9, a_{10} be an arithmetic sequence.
If a_1 + a_3 + a_5 = 5 and a_2 + a_4 = -2, then find a_1.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        As this problem is posed in the post,  it is  FATALLY  WRONG,
        and I will explain below,  WHY  it is so. 


If  the terms  a_1, a_2, a_3, a_4, a_5  form arithmetic sequence with the common difference d, 

then the terms  a_1, a_3, a_5  also form arithmetic sequence, this time with the common difference 2d.


We can write these two equalities

    a_1 = a_3 - 2d,

    a_5 = a_3 + 2d.


By adding them, we get  a_1 + a_5 = 2a_3.

By adding a_3 to it, we get

    a_1 + a_3 + a_5 = 3a_3.


So,  3a_3 = 5 ,  and we get  a_3 = 5/3.



Similarly, we can write these two equalities

    a_2 = a_3 - d,

    a_4 = a_3 + d.


By adding them, we get

    a_2 + a_4 = 2a_3,

or

    2a_3 = -2,  a_3 = -2/2 = -1.


Thus we get for a_3 two different values, 5/3 and -1  simultaneously,   which is impossible and NEVER may happen.


So, we conclude that the problem is FATALLY INCONSISTENT.


It is SELF-CONTRADICTORY and describes a situation, which never may happen.

From it,  I make a conclusion,  that a person,  who created this  " problem ",  is mathematically illiterate,
does not know the subject,  produces  HIBBERISH  and disseminates it in the  Internet,
confusing  readers and making them to waste their precious time,  which is  REGRETTABLE.


/\/\/\/\/\/\/\/\/\/\/\/

        To the managers of this project


As I explained in my post,  a person,  who created this  " problem ",  is mathematically illiterate,
does not know the subject,  produces  HIBBERISH  and disseminates it in the  Internet,
confusing  readers and making them to waste their precious time,  which is  REGRETTABLE.


                        Please take the measures.



Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

I agree with the other tutor. The way this problem is set up leads to a contradiction.
a1+a3+a5 = 5
a2+a4 = -2

Add the equations to arrive at
a1+a2+a3+a4+a5 = 3
which means
S5 = 3

S5 represents the sum of the first 5 terms of the arithmetic sequence.

Sn = sum of the first n terms of an arithmetic sequence
Sn = (n/2)*(a1 + an)
S5 = (5/2)*(a1 + a5)
3 = 2.5*(a1 + a5)
a1 + a5 = 3/(2.5)
a1 + a5 = 1.2

a1+a3+a5 = 5
(a1+a5)+a3 = 5
1.2+a3 = 5
a3 = 5-1.2
a3 = 3.8

--------------------------------------------------------------------------

Then consider:
a2+a4 = -2
(a3-d)+(a3+d) = -2
2*a3 = -2
a3 = -2/2
a3 = -1

But this contradicts a3 = 3.8 found in the previous section.
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


Since the sequence a_1, a_2, a_3, ..., a_10 is arithmetic, the sequences

[1] a_1, a_3, a_5

and

[2] a_2, a_4

are also both arithmetic.

In [1], a_3 is the middle term of the three, so it is the average of the three terms; since a_1 + a_3 + a_5 = 5, a_3 must be 5/3.

But in [2], a_3 is halfway between a_2 and a_4, so a_3 must be the average of a_2 and a_4. And a_2 + a_4 = -2, which means a_3 must be -2/2 = -1.

So the conditions of the problem require that a_3 be both 5/3 and -1, which is a contradiction.

ANSWER: The problem is faulty
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