Question

Find the $8^{th}$ term of the sequence whose first three terms are $3,3,6$ and each term after the second is the sum of three two terms preceding it .

Answer 1

Given $3,3,6$………

Now $3,3,6,9,15,24,39,63,102$

(It involves smaller numbers)

$3$

$3+0=3$

$3+3=6$

$3+6=9$

$6+9=15$

$9+15=24$

$15+24=39$

$24+39=63$ ($8^{th}$ term)

$\therefore 63$ is $8^{th}$ term.