Solve previous year Question of iit jee
Chapter :- sequence and series
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7
Answer:
a, b , c are in AP
Step-by-step explanation:
Given :- 25a² + 9b² + 25c² - 7ac - 45ab - 15bc = 0
→ 225a² + (3b)² + 5c² - (15a)(15c) - (15a )(3b) - (3b) (5c) = 0
→ 1/2 [(15a-3b)² + (3b-5c²) + (5c-15a)² ] =0
→ 1/2 [(15a - 3b)² + (3b - 5c)² + (5c - 15a)² ] = 0
→ 15a = 3b , 3b= 5c , and 5c = 15a
→ 15a = 3b = 5c
→ a/1 = b/5 = c/3 = λ (assume)
→ a = λ , b= 5λ , c = 3λ
Ηence, a, b and c are in A.P
Answered by
6
Answer:
Given :-
25a² + 9b² + 25c² - 7ac - 45ab - 15bc = 0
→ 225a² + (3b)² + 5c² - (15a)(15c) - (15a )(3b) - (3b) (5c) = 0
→ 1/2 [(15a-3b)² + (3b-5c²) + (5c-15a)² ] =0
→ 1/2 [(15a - 3b)² + (3b - 5c)² + (5c - 15a)² ] = 0
→ 15a = 3b , 3b= 5c , and 5c = 15a
→ 15a = 3b = 5c
→ a/1 = b/5 = c/3 = λ (assume)
→ a = λ , b= 5λ , c = 3λ
Ηence, a, b and c are in A.P
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