Question

For a sequence,if so =2(3raised to n-1, find the nth term hence show that sequence is a GP

Answer 1

Answer:

will try to answer your question please mark me as Brainliest and follow me❤

Step-by-step explanation:

❤❤.........

Answer 2

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

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 2