Question 958856
The product of the first and the third terms of an arithmetic sequence is 5. If
all terms of the sequence are positive integers, what is the fourth term?
<pre>
{{{a[1]*a[3]=5}}}
{{{a[3]=a[1]+(3-1)d}}}
{{{a[3]=a[1]+2d}}}

{{{a[1]*a[3]=5}}}

{{{a[1](a[1]+2d)=5}}}

{{{(a[1])^2 +2d*a[1]-5=0

{{{a[1] = (-2d +- sqrt( (2d)^2-4*1*(-5) ))/(2*1) }}}

{{{a[1] = (-2d +- sqrt(4d^2+20))/2 }}}

{{{a[1] = (-2d +- sqrt(4(d^2+5)))/2 }}}

{{{a[1] = (-2d +- 2sqrt(d^2+5))/2 }}}

{{{a[1] = (2(-d +- sqrt(d^2+5)))/2 }}}

{{{a[1] = -d +- sqrt(d^2+5) }}}

What's under the square root must be a perfect square
since {{{a[1]}}} is a positive integer. The first
perfect square greater than 5 is 9, and what's under
the square root will be 9 if d=2

Substituting d=2 in

{{{a[1] = -d +- sqrt(d^2+5) }}}

{{{a[1] = -2 +- sqrt(2^2+5) }}}

{{{a[1] = -2 +- sqrt(9) }}}

{{{a[1] = -2 +- 3 }}}

We must take the + since {{{a[1]}}} is a positive integer.

{{{a[1] = -2 + 3 }}}

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

Since d = 2, then the sequence is

1,3,5,7

The 4th term is 7.

Edwin</pre>