### Math Notes

Subjects

#### Algebra Solutions

##### Topics || Problems

If the 6th term of a geometric progression (g.p.) is 4 and the 10th term is 4/81, find r.

$$a_n = a_1 r^{n-1}$$

$$a_6 = a_1 r^{6-1}$$

$$4 = a_1 r^5$$

$$a_40 = a_1 r^{10-1}$$

$$a_10 = a_1 r^{10-1}$$

$$\frac{4}{81} = a_1 r^9$$

Divide the two equations to eliminate $$a_1$$

$$\frac{4}{81(4)} = \frac{a_1r^9}{a_1r^5}$$

$$\frac{1}{81} = r^4$$

$$r=\frac{1}{3}$$ Answer