Question

G.P.: \[\frac{2}{9}, \frac{1}{3}, \frac{1}{2}, \ldots\]. 

Assertion (A): The \[5^{\text{th}}\] term of the given G.P. is \[1\frac{1}{8}\].

Reason (R): If for a G.P., the first term is a, the common ratio is r and the number of terms is n, then the sum of the first n terms is \[S_{n}=\frac{a(r^{n}-1)}{r-1}\] for all r.

Options

• A is true, R is false.

• A is false, R is true.

• Both A and R are true and R is the correct reason for A.

• Both A and R are true and R is the incorrect reason for A.

Answer 1

A is true, R is false.

Explanation:

Given, the sequence \[{} = \frac{2}{9}, \frac{1}{3}, \frac{1}{2}, .............\]

First term (a) \[{} = \frac{2}{9}\]

Common ratio (r) \[{} = \frac{\frac{1}{3}}{\frac{2}{9}} = \frac{1 \times 9}{2 \times 3} = \frac{3}{2}\]

Using the formula; \[T_n = a \cdot r^{n \: - \: 1}\]

\[T_5 = \frac{2}{9} \times \left(\frac{3}{2}\right)^{5 \: - \: 1}\]

\[{} = \frac{2}{9} \times \left(\frac{3}{2}\right)^4\]

\[{} = \frac{2}{9} \times \frac{81}{16}\]

\[{} = \frac{9}{8}\]

\[{} = 1\frac{1}{8}.\]

So, assertion (A) is true.

According to reason:

Sum of first \[n\] terms \[(S_n) = \frac{a(r^n - 1)}{r - 1}\], for all \[r.\]

But this is not the case when \[r = 1.\]

Thus,

Reason (R) is false.