If the first three Fibonacci nos. are x1=1, x2=1 and x3=2, what is the value of n for which xn + xn+1 = 89 ?

11

Step-by-step explanation:

Looking at Fibonacci numbers ;

We have:

Than X1 =1 ; X2 = 1; X3 = 2

Xn+1 = Xn + Xn-1

X3 = X2 + X1 = 1+1 =2; X4 = X3 + X2 = 2 +1 = 3; X5= X4+X3 = 3+2 = 5; X6 = X5 + X4 = 5 + 3 = 8; X7 = X6 +X5 = 8 +5 = 13; X8 = X7 + X6 = 13 + 8 = 21; X9 = X8 + X7 = 21+ 13 = 34 ; X10 = X9 + X8 = 34 +21 = 55 ; X11 = X10 + X9 = 55 +34 = 89

We see that ; X11 gives 89 so the value of n = 11

x² + 11x + 18 < 0  

[-11 ± sqrt(121 - 72)]/2 = (-11 ± 7)/2  

solution: (-9, -2)

step-by-step explanation: