CUET UG 2025 — General Test Quantitative Reasoning

The series is:

$$1 + \frac{3}{2} + 2 + \frac{5}{2} + 3 + \frac{7}{2} + ...$$1+23​+2+25​+3+27​+...

Converting to fractions:

$$\frac{2}{2} + \frac{3}{2} + \frac{4}{2} + \frac{5}{2} + \frac{6}{2} + \frac{7}{2} + ...$$22​+23​+24​+25​+26​+27​+...

The numerators are 2, 3, 4, 5, 6, 7... (consecutive numbers starting from 2)

---

For the nth term:

• 1st term has numerator = 2 = (1+1)
• 2nd term has numerator = 3 = (2+1)
• 3rd term has numerator = 4 = (3+1)

General term: $T_n = \frac{n+1}{2}$Tn​=2n+1​

---

The sum of n terms:

$$S_n = \sum_{k=1}^{n} \frac{k+1}{2}$$Sn​=k=1∑n​2k+1​

$$S_n = \frac{1}{2} \sum_{k=1}^{n} (k+1)$$Sn​=21​k=1∑n​(k+1)

$$S_n = \frac{1}{2} \left[\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1\right]$$Sn​=21​[k=1∑n​k+k=1∑n​1]

Using $\displaystyle \sum_{k=1}^{n} k = \frac{n(n+1)}{2}$k=1∑n​k=2n(n+1)​ and $\displaystyle \sum_{k=1}^{n} 1 = n$k=1∑n​1=n:

$$S_n = \frac{1}{2} \left[\frac{n(n+1)}{2} + n\right]$$Sn​=21​[2n(n+1)​+n]

---

$$S_n = \frac{1}{2} \left[\frac{n(n+1) + 2n}{2}\right]$$Sn​=21​[2n(n+1)+2n​]

$$S_n = \frac{1}{2} \left[\frac{n(n+1+2)}{2}\right]$$Sn​=21​[2n(n+1+2)​]

$$S_n = \frac{n(n+3)}{4}$$Sn​=4n(n+3)​

Therefore, the sum of n terms is $\frac{n(n+3)}{4}$4n(n+3)​