115 + 135 + 163 + … + 1195 = ?
Answer और ExplanationAnswer and explanation
Answer: (B) 2/15
हरों (Denominators) को गुणनखंडों (Factors) में तोड़ें:
15 = 3 × 5, 35 = 5 × 7, 63 = 7 × 9, …, 195 = 13 × 15
हर भिन्न (Fraction) 1k(k + 2) = 12(1k − 1k + 2) है
योग = 12[(13 − 15) + (15 − 17) + … + (113 − 115)]
= 12(13 − 115)
= 12 × 415 = 215
✓
योग 215 है।
Break the denominators into factors:
15 = 3 × 5, 35 = 5 × 7, 63 = 7 × 9, …, 195 = 13 × 15
Each fraction is 1k(k + 2) = 12(1k − 1k + 2)
Sum = 12[(13 − 15) + (15 − 17) + … + (113 − 115)]
= 12(13 − 115)
= 12 × 415 = 215
✓
The sum is 215.
SMART APPROACH
d = 2, पहली संख्या 3, आखिरी संख्या 15: 12(13 − 115) = 12 × 415 = 215।
d = 2, first number 3, last number 15: 12(13 − 115) = 12 × 415 = 215.
Source: MTS · 05 Feb 2025 · Shift 1। Solution: Prashant Sir, Vijayastra।Source: MTS · 05 Feb 2025 · Shift 1. Solution by Prashant Sir, from Vijayastra.