For a sequence,if so =2(3raised to n-1, find the nth term hence show that sequence is a GP
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Answer:
Answer-
Answer-( identity : (a-b)² = a²-2ab +b²)
Answer-( identity : (a-b)² = a²-2ab +b²)= {(2x - 5y)}^{2}=(2x−5y)
Answer-( identity : (a-b)² = a²-2ab +b²)= {(2x - 5y)}^{2}=(2x−5y) 2
Answer-( identity : (a-b)² = a²-2ab +b²)= {(2x - 5y)}^{2}=(2x−5y) 2
Answer-( identity : (a-b)² = a²-2ab +b²)= {(2x - 5y)}^{2}=(2x−5y) 2 = {(2x)}^{2} - 2(2x)(5y) + {(5y)}^{2}=(2x)
Answer-( identity : (a-b)² = a²-2ab +b²)= {(2x - 5y)}^{2}=(2x−5y) 2 = {(2x)}^{2} - 2(2x)(5y) + {(5y)}^{2}=(2x) 2
Answer-( identity : (a-b)² = a²-2ab +b²)= {(2x - 5y)}^{2}=(2x−5y) 2 = {(2x)}^{2} - 2(2x)(5y) + {(5y)}^{2}=(2x) 2 −2(2x)(5y)+(5y)
Answer-( identity : (a-b)² = a²-2ab +b²)= {(2x - 5y)}^{2}=(2x−5y) 2 = {(2x)}^{2} - 2(2x)(5y) + {(5y)}^{2}=(2x) 2 −2(2x)(5y)+(5y) 2
Answer-( identity : (a-b)² = a²-2ab +b²)= {(2x - 5y)}^{2}=(2x−5y) 2 = {(2x)}^{2} - 2(2x)(5y) + {(5y)}^{2}=(2x) 2 −2(2x)(5y)+(5y) 2
Answer-( identity : (a-b)² = a²-2ab +b²)= {(2x - 5y)}^{2}=(2x−5y) 2 = {(2x)}^{2} - 2(2x)(5y) + {(5y)}^{2}=(2x) 2 −2(2x)(5y)+(5y) 2 = 4 {x}^{2} - 20xy + 25 {y}^{2}=4x
Answer-( identity : (a-b)² = a²-2ab +b²)= {(2x - 5y)}^{2}=(2x−5y) 2 = {(2x)}^{2} - 2(2x)(5y) + {(5y)}^{2}=(2x) 2 −2(2x)(5y)+(5y) 2 = 4 {x}^{2} - 20xy + 25 {y}^{2}=4x 2
Answer-( identity : (a-b)² = a²-2ab +b²)= {(2x - 5y)}^{2}=(2x−5y) 2 = {(2x)}^{2} - 2(2x)(5y) + {(5y)}^{2}=(2x) 2 −2(2x)(5y)+(5y) 2 = 4 {x}^{2} - 20xy + 25 {y}^{2}=4x 2 −20xy+25y