UPSC NDA II 2023 Mathematics Solved Paper

Concept:In an arithmetic progression (AP), the sum of selected terms can be expressed in terms of the first term $a$ and common difference $d$.The sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$.Explanation:Let the first term be $a$ and the common difference be $d$.The $n$-th term is $a_n = a + (n-1)d$.The given condition is $a_1 + a_5 + a_{10} + a_{15} + a_{20} + a_{25} + a_{30} + a_{34} = 300$.Substitute each term: $a + (a+4d) + (a+9d) + (a+14d) + (a+19d) + (a+24d) + (a+29d) + (a+33d) = 300$.This simplifies to $8a + 132d = 300$.Divide both sides by $4$ to get $2a + 33d = 75$.We need $\sum_{n=1}^{34} a_n = S_{34}$.Using the sum formula: $S_{34} = \frac{34}{2}[2a + (34-1)d] = 17[2a + 33d]$.Substitute $2a + 33d = 75$: $S_{34} = 17 \times 75 = 1275$.Answer:$\sum_{n=1}^{34} a_n = 1275$, which corresponds to option D.