The eleventh term in the following sequence: -3/2,3/4,3/8,3/16,-3/32,… is :

Correct Answer is : −32048-3/2048−20483

Correct Answer is : −32048-3/2048−20483

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UGC NET Paper I 8 Jan 2025 Shift 1 Paper

Question : 25 of 50

Marks: +1, -0

-\frac{3}{2},\frac{3}{4},\frac{3}{8},\frac{3}{16},-\frac{3}{32},…

is :

Solution:

Concept:

The sequence is a geometric progression where each term is obtained by multiplying the previous term by a fixed common ratio.

Explanation:

First term

a = -\frac{3}{2}

Common ratio

r

is found by dividing the second term by the first:

\frac{3}{4} \div -\frac{3}{2} = \frac{3}{4} \times -\frac{2}{3} = -\frac{1}{2}

The general term of a geometric sequence is

T_n = a \times r^{(n-1)}

For the 11th term,

n = 11

T_{11} = -\frac{3}{2} \times (-\frac{1}{2})^{10}

Since 10 is an even power,

(-\frac{1}{2})^{10} = \frac{1}{2^{10}} = \frac{1}{1024}

Therefore,

T_{11} = -\frac{3}{2} \times \frac{1}{1024} = -\frac{3}{2048}

Answer:

-\frac{3}{2048}

(Option D).

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