IBPS PO Mains 12 Oct 2025 Shift 1 Paper

Concept:Identify the pattern in Series A to find $P$, then use the pattern of decreasing even-number squares to complete Series B and compute $S$.Explanation:In Series A, each term from the 3rd term onward is obtained by multiplying the previous term by an increasing multiplier and adding 1.$3 \times 5 + 1 = 16$$16 \times 6 + 1 = 97$$97 \times 7 + 1 = 680$$680 \times 8 + 1 = 5441$Hence, $P = 680$.Now find the 3rd term of Series B using $P$.$3.5 \times 680 + 172 = 2380 + 172 = 2552$So Series B has the known terms: $Q$, $Q-400$, $2552$, $2296$, $2100$, $1956$, $R$.Check the differences between consecutive known terms.$2552 - 2296 = 256 = 16^2$$2296 - 2100 = 196 = 14^2$$2100 - 1956 = 144 = 12^2$Thus, the differences are squares of even numbers in descending order: $20^2, 18^2, 16^2, 14^2, 12^2, 10^2$.Find the missing terms using these differences.$R = 1956 - 10^2 = 1956 - 100 = 1856$$Q - 400 = 2552 + 18^2 = 2552 + 324 = 2876$$Q = 2876 + 400 = 3276$Therefore, Series B is: $3276, 2876, 2552, 2296, 2100, 1956, 1856$.$S = 3276 + 2876 + 2552 + 2296 + 2100 + 1956 + 1856 = 16912$$P \times 2 = 680 \times 2 = 1360$$S - (P \times 2) = 16912 - 1360 = 15552$Answer:$15552$ (Option D).