Check whether the following sequences are geometric progressions and find their nth terms, Class 9

Check whether the following sequences are geometric progressions and find their n^th terms, Class 9

Question 1.

Check whether the following sequences are geometric progressions and find their n^th terms, Class 9

i. 2, 10, 50, 250,...

ii. $4, \frac{8}{3}, \frac{16}{9}, \frac{32}{27} \ldots$

iii. $3,-\frac{3}{2}, \frac{3}{4},-\frac{3}{8}, \ldots$

Answer:

To check that sequences are in geometric Progressions, we have to find the ratio between the consecutive terms:

i. Given sequence is 2,10, 50, 250....

$\therefore \quad \frac{10}{2}=5, \frac{50}{10}=5, \frac{250}{50}=5$

Since the ratio between two terms is constant. The given sequence is a G.P.

First term (a) = 2 and common ratio (r) = 5

∴ n^th term: t_n = ar^n-1

∴ t_n = 2(5)^n-1

ii. Given sequence is $4, \frac{8}{3}, \frac{16}{9}, \frac{32}{27} \ldots$

$\therefore \quad \frac{\frac{8}{3}}{4}=\frac{2}{3}, \frac{\frac{16}{9}}{\frac{8}{3}}=\frac{2}{3}$

Since the ratio between two terms is constant. The given sequence is a G.P.

First term (a) = 4 and

common ratio (r) = $\frac{2}{3}$

∴ n^th term: t_n = ar^n-1

∴ t_n = 4$4\left(\frac{2}{3}\right)^{n-1}$

iii. Given sequence is $3,-\frac{3}{2}, \frac{3}{4}, \frac{-3}{8} \ldots$

$\therefore \quad \frac{-\frac{3}{2}}{3}=-\frac{1}{2}, \frac{\frac{3}{4}}{-\frac{3}{2}}=-\frac{1}{2}$

Since the ratio between two terms is constant. The given sequence is a G.P. First term (a) = 3 and common ratio (r) = $-\frac{1}{2}$

∴ n^th term: t_n = ar^n-1

∴ t_n = $3\left(-\frac{1}{2}\right)^{n-1}$