Question #79b63

Question

1599 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-07T15:18:37

1st term=3
2nd term =12
3rd term =48

Let
$the 1st term of GP = a$ $" the 1st term of GP "=a$
$and common ratio = r$ $" and common ratio "=r$

$So 2nd term = a r and 3rd term = a r^{2}$ $" So 2nd term"= ar " and 3rd term"=ar^2$

By the 1st condition

$\frac{a r^{2}}{a} = 16 \Rightarrow r = \sqrt{16} = 4$ $(ar^2)/a=16=>r=sqrt16=4$

By the 2nd condition

$a r^{2} - a = 45 \Rightarrow a (4^{2} - 1) = 45 \Rightarrow a = 3$ $ar^2-a=45=>a(4^2-1)=45=>a=3$

Hence

$1st term = 3$ $"1st term "=3$

$2nd term = 12$ $"2nd term "=12$

$3rd term = 48$ $"3rd term "= 48$