Do you think this is Fibonacci?

Assume the first two terms to be $x$ and $y$ respectively. Then, we can construct the six member series as:

$x,y,x+y,x+2y,2x+3y,3x+5y$

It's given that the last term is $4$ times the first, hence it follows that:

$3x+5y=4x \Rightarrow 5y=x$ ............................................. $(1)$

Also, sum of all terms is $13$ , hence:

$8x+12y=13$ ........................................... $(2)$

Solving $(1)$ and $(2)$ , we get:

$x=5/4$ and $y=1/4$

Hence the second term is $\boxed{1/4}$

let first no. be "x"

let second no. be "y"

the sequence would be:

(x)+(y)+(x+y)+(x+2y)+(2x+3y)+(3x+5y)=13

8x+12y=13 ......(i)

Also,

4x=3x+5y

x=5y .....(ii)

by substituting (ii) in (i)

8(5y)+12y=13

40y+12y=13

52y=13

y=1/4

Using the recurrence relation: $F_{6}$ = $F_{5}$ + $F_{4}$ = 2 $F_{4}$ + $F_{3}$ =...= 5 $F_{2}$ + 3 $F_{1}$ . Now we can see that the coefficients of $F_{2}$ and $F_{1}$ are the 5th and 4th terms of the Fibonacci sequence (respectively). Using this fact produces: $F_{5}$ = 3 $F_{2}$ + 2 $F_{1}$ , $F_{4}$ = 2 $F_{2}$ + $F_{1}$ , $F_{3}$ = $F_{2}$ + $F_{1}$ . Now summing these terms together with $F_{2}$ and $F_{1}$ , gives: 12 $F_{2}$ + 8 $F_{1}$ = 13 (#1). Now using the recurrence relation with [ $F_{6}$ = 4 $F_{1}$ ] $\rightarrow$ $F_{6}$ + $F_{5}$ + ... + $F_{1}$ = ( $F_{5}$ + $F_{4}$ + $F_{3}$ + $F_{2}$ + 5 $F_{1}$ ) = ( $F_{6}$ + $F_{3}$ + $F_{2}$ + 5 $F_{1}$ ) = ( $F_{3}$ + $F_{2}$ + 9 $F_{1}$ ). Using $F_{1}$ = $F_{3}$ - $F_{2}$ and - $F_{2}$ = $F_{3}$ - $F_{4}$ $\rightarrow$ 13 = $F_{4}$ +[ $F_{3}$ - $F_{2}$ ] + 8 $F_{1}$ = [ $F_{4}$ + $F_{3}$ ] + [ $F_{3}$ - $F_{4}$ ] + 8 $F_{1}$ = 2 $F_{3}$ + 8 $F_{1}$ = 2 $F_{2}$ +10 $F_{1}$ $\Rightarrow$ $F_{2}$ + 5 $F_{1}$ = $\frac{13}{2}$ (#2). Now let ${x}$ = $F_{2}$ and $F_{1}$ = ${y}$ , such that (from #1 and #2): 3 ${x}$ + 2 ${y}$ = $\frac{13}{4}$ and ${x}$ + 5 ${y}$ = $\frac{13}{2}$ . This solves to give: ${x}$ = $\boxed{\frac{1}{4}}$ and ${y}$ = $\frac{5}{4}$ .

• Assume that the first and second term is a and b, so we get

a, b, a+b, a+2b, 2a+3b, 3a+5b

• The last term is 4 times the first, means that

3a+5b=4a or a=5b

• the sum of all terms is 13

a+b+(a+b)+(a+2b)+(2a+3b)+(3a+5b)=13

8a+12b=13

8(5b)+12b=13

52b=13

b=1/4

• Assuming the first two terms to be x and y , respectively, we have the six sequence terms are:
  I : {x II: {y II: {x + y IV: {x + 2y V: {2x + 3y VI: {3x + 5y .
  
• Now, we have this system of equations: {3x + 5y = 4x {8x + 12y = 13
• Developing the first equation, we find $x = 5y$ .
• Replacing it in the second equation, we'll have:
  $40y + 12y = 13 => 52y = 13$
  
• Concluding, we have that the second term, (which is fortunatly y) as $y = \frac{1}{4}$