Direction : Study the following information carefully and answer the given question. Observe the Series I carefully to identify the logic and obtain the value of 'P'. In both of the series, the same logic is applied. Series I: 60, 68, 14, (P), -294, 786, -1272 Series II: (P-12), Q, R, S, T, U, W What is the value of 'R'?

### Concept & Number Series Logic To solve this dual number series, we must first determine the mathematical pattern governing Series I. By analyzing the differences between consecutive terms, we can identify a pattern involving cubes, multipliers, and alternating signs. $$ T_n = T_{n-1} + (-1)^n \times (n-1) \times n^3 $$ (Where $n$ represents the term's positional index in the series, starting from $n=2$ for the first difference.) ### Step-by-Step Solution 1. **Analyze Series I:** * $T_1 = 60$ * $T_2 = 68 \rightarrow 60 + 8 \rightarrow 60 + (1 \times 2^3)$ * $T_3 = 14 \rightarrow 68 - 54 \rightarrow 68 - (2 \times 3^3)$ * $T_4 (P) \rightarrow 14 + (3 \times 4^3) \rightarrow 14 + (3 \times 64) = 14 + 192 = 206$ * Let's verify the pattern for the next term: * $T_5 = 206 - (4 \times 5^3) \rightarrow 206 - (4 \times 125) = 206 - 500 = -294$ (Matches exactly) * Hence, **$P = 206$**. 2. **Evaluate Series II up to 'R':** The logic remains identical. * First Term ($T_1$) = $P - 12 = 206 - 12 = 194$ * Second Term ($Q$) = $194 + (1 \times 2^3) = 194 + 8 = 202$ * Third Term ($R$) = $202 - (2 \times 3^3) = 202 - 54 = 148$ ### Exam Strategy & Shortcut When you see numbers dropping and rising sharply (like 68 to 14, then up to a large positive, then down to -294), immediately suspect alternating addition/subtraction combined with squares or cubes. Finding the differences ($+8, -54$) quickly reveals $2^3$ and a multiple of $3^3$. ### Common Pitfall A common mistake is failing to recognize the increasing multiplier ($1, 2, 3, \dots$) alongside the cubes, leading to incorrect projections for $P$ and subsequent terms in Series II. ### Final Answer Therefore, the correct answer is **148**.