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UGC-NET-Paper1

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What will be the eleventh term in the following sequence?
-3/2, 3/4, -3/8, 3/16, -3/32, ......

1. 3/1024
2. -3/1024
3. 3/2048
4. -3/2048

निम्नलिखित क्रम में ग्यारहबाँ पद क्या होगा ?
-3/2, 3/4, -3/8, 3/16, -3/32, ......

1. 3/1024
2. -3/1024
3. 3/2048
4. -3/2048
This Question came in
UGC-NET-Hindi-08-January-2025-Shift-1-Q37
UGC-NET Paper 1 Full Course

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Detailed Explanation & Answer
The signs alternate: negative, positive, negative, positive, ...

So, odd terms are negative, even terms are positive.
Since 11 is odd, the 11th term will be negative.

Step 1: Denominator pattern
2, 4, 8, 16, 32, ... → This is a geometric progression:
2 = 2¹
4 = 2²
8 = 2³
...
So, the nth term has denominator = 2ⁿ

For the 11th term: denominator = 2¹¹ = 2048

Step 2: Numerator pattern
Numerator stays constant: always 3, alternating signs.

So, 11th term = -3/2048
चिह्न एक दूसरे से बदलते हैं: ऋणात्मक, धनात्मक, ऋणात्मक, धनात्मक, ...

इसलिए, विषम पद ऋणात्मक हैं, सम पद धनात्मक हैं।

चूँकि 11 विषम है, इसलिए 11वाँ पद ऋणात्मक होगा।

चरण 1: हर पैटर्न
2, 4, 8, 16, 32, ... → यह एक ज्यामितीय प्रगति है:
2 = 2¹
4 = 2²
8 = 2³
...
इसलिए, nवें पद का हर = 2ⁿ है

11वें पद के लिए: हर = 2¹¹ = 2048

चरण 2: अंश पैटर्न
अंश स्थिर रहता है: हमेशा 3, चिह्न बदलते रहते हैं।

इसलिए, 11वाँ पद = -3/2048
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