CUET UG 2025 — General Test Quantitative Reasoning

The first term is $a_1 = 7$a1​=7 and the 6th term is $a_6 = 22$a6​=22.

The general term of an arithmetic progression is:

$$a_n = a_1 + (n - 1) \times d$$an​=a1​+(n−1)×d

For the 6th term:

$$a_6 = a_1 + (6 - 1) \times d$$a6​=a1​+(6−1)×d

$$22 = 7 + 5d$$22=7+5d

$$15 = 5d$$15=5d

$$d = 3$$d=3

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The sum of the first $n$n terms of an arithmetic progression is:

$$S_n = \frac{n}{2} \times [2a_1 + (n - 1) \times d]$$Sn​=2n​×[2a1​+(n−1)×d]

For the first 10 terms with $a_1 = 7$a1​=7 and $d = 3$d=3:

$$S_{10} = \frac{10}{2} \times [2(7) + (10 - 1) \times 3]$$S10​=210​×[2(7)+(10−1)×3]

$$S_{10} = 5 \times [14 + 9 \times 3]$$S10​=5×[14+9×3]

$$S_{10} = 5 \times [14 + 27]$$S10​=5×[14+27]

$$S_{10} = 5 \times 41$$S10​=5×41

$$S_{10} = 205$$S10​=205

Therefore, the sum of the first 10 terms is $205$205.