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In the sequence of 3 consecutive natural number, product of two alternate number is one less than the square of middle number.1×3=2skoyar -18×10=9skoyar -1using this, finda)99×101=------b)49×51=----c)199×201=---- |
Answer» SOLUTIONGIVEN In the SEQUENCE of 3 consecutive NATURAL number, product of two alternate number is ONE less than the square of middle number. 1 × 3 = 2² - 1 8 × 10 = 9² -1 TO DETERMINE Using this, find a) 99×101 = b) 49×51 = c) 199×201 = EVALUATION Here it is given that In the sequence of 3 consecutive natural number, product of two alternate number is one less than the square of middle number. 1 × 3 = 2² - 1 8 × 10 = 9² -1 Now a) 99×101 = 100² - 1 = 10000 - 1 = 9999 b) 49×51 = 50² - 1 = 2500 - 1 = 2499 c) 199×201 = 200² - 1 = 40000 - 1 = 39999 ━━━━━━━━━━━━━━━━ Learn more from Brainly :-
2. to verify algebraic IDENTITY a2-b2=(a+b)(a-b) |
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