Question

Find six numbers in A.P., such that the sum of two extremes be 16 and the product of the two middle terms be 63.

Answer 1

If the sum of the first and last numbers is an integer and the product of the two middle terms is an integer, then almost certainly all the numbers will be integers.

Then if the product of the middle two numbers is 63, the two middle numbers are 7 and 9.

So the common difference is 2; that makes the 6 numbers

3, 5, 7, 9, 11, 13


You could, of course, solve the problem using algebra; however, it takes far more work than the logical analysis solution shown above.

Let the first term be a and the common difference be d; then the 6 numbers are
a, a+d, a+2d, a+3d, a+4d, a+5d

Then
(1) 2a%2B5d+=+16 the sum of the first and last numbers is 16
(2) %28a%2B2d%29%28a%2B3d%29+=+63 the product of the two middle numbers is 63

Solve equation (1) for a: a+=+%2816-5d%29%2F2

Substitute into equation (2):
%28%2816-d%29%2F2%29%28%2816%2Bd%29%2F2%29+=+63
%28256-d%5E2%29%2F4+=+63
256-d%5E2+=+252
d%5E2+=+4
d+=+2

Substitute back into equation (1) to find a:
2a%2B10+=+16
2a+=+6
a+=+3


I think I like the logical analysis solution better....

Answer by ikleyn(20438) (Show Source): You can put this solution on YOUR website!
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Find six numbers in AP, such that the sum of the two extremes be 16 and the product of the two middle terms be 63.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

If the sum of the two extreme terms of the AP is 16, then the sum of any two other terms, equally remoted from the end-terms is 16, too.


It is WELL KNOWN property of any AP, the property which each student should learn when he or she studies the sum of an arithmetic progression.


Hence, in our case, if a%5B1%5D%2Ba%5B6%5D = 16, then

a%5B2%5D%2Ba%5B5%5D = 16,

a%5B3%5D%2Ba%5B4%5D = 16.


Thus we can reformulate the original problem in this way:


Find six numbers in AP, such that the sum of the two middle terms be 16 and the product of the two middle terms be 63.


So, the key moment is to find two number whose sum is 16 and the product is 63.


It is very elementary and standard problem. You can solve it using quadratic equation.

But it admits absolutely elementary MENTAL solution: 63 = 7*9 and 7 + 9 = 16.


Thus we found that a%5B3%5D = 7 and a%5B4%5D = 9.


Then the common difference is 9-7 = 2, and the AP is, obviously

3, 5, 7, 9, 11, 13.


By the way, the reversed progression

13, 11, 9, 7, 5, 3

is the solution, too.

SOLVED

Answer 2

Answer:

The six numbers in A.P  are $3, 5, 7, 9, 11, 13.$

Step-by-step explanation:

Step 1

If the totality of the first and last numbers exists as an integer and the product of the two middle terms exists as an integer, then almost certainly all the digits will be integers.

Then if the product of the middle two numbers exists $63$, the two middle numbers are $7$ and $9.$

So the common difference exists $2$; that generates the $6$ digits

$3, 5, 7, 9, 11, 13$

Using algebra; however, it takes distant more work than the logical analysis solution shown above.

Step 2

Let, first the term be the 'a' and the common difference be 'd'.

$a, a+d, a+2d, a+3d, a+4d, a+5d$

Then

$(1)$ $2 a+5 d=16$ the sum of the first and last numbers is $16$

$(2)$ $(a+2 d) \cdot(a+3 d)=63$ the product of the two middle numbers is $63$  

Step 3

Solve equation (1) for $a$ :

$a=\frac{16-5 d}{2}$

Substitute into equation (2):

$\left(\frac{16-d}{2}\right) \cdot\left(\frac{16+d}{2}\right)=63$

$&\frac{256-d^{2}}{4}=63$

$&256-d^{2}=252$

$&d^{2}=4$

$&d=2$

Substitute this in the equation $(1),$

$&2 a+10=16$

$&2 a=6$

$&a=3$

Therefore, the six numbers in A.P $3, 5, 7, 9, 11, 13.$

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