CUET UG 2025 — General Test Quantitative Reasoning

An arithmetic progression has the formula:

$$a_n = a + (n - 1)d$$an​=a+(n−1)d

Where $a$a is the first term, $n$n is the position, and $d$d is the common difference.

Given that the 21st term is 91 and the 33rd term is 145.

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Using the formula for both given terms:

$a_{21} = a + 20d = 91$a21​=a+20d=91 ... ①

$a_{33} = a + 32d = 145$a33​=a+32d=145 ... ②

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Subtracting equation ① from equation ②:

$$(a + 32d) - (a + 20d) = 145 - 91$$(a+32d)−(a+20d)=145−91

$$12d = 54$$12d=54

$$d = \frac{54}{12}$$d=1254​

$$d = 4.5$$d=4.5

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Substituting $d = 4.5$d=4.5 into equation ①:

$$a + 20(4.5) = 91$$a+20(4.5)=91

$$a + 90 = 91$$a+90=91

$$a = 1$$a=1

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Finding the 29th term:

$$a_{29} = a + (29 - 1)d$$a29​=a+(29−1)d

$$a_{29} = 1 + 28(4.5)$$a29​=1+28(4.5)

$$a_{29} = 1 + 126$$a29​=1+126

$$a_{29} = 127$$a29​=127

Therefore, the 29th term is 127.

The answer is option 1.