Question

pls refer to the attachment​

Answer 1

Answer:-

Given:

• a , b , c -- AP

As we know, d (Common difference) = b - a (or) c - b

b - a = c - b

=> a - b = b - c -- equation (1)

And,

- b - b = - c - a

2b = a + c -- equation (2)

• x , y , z -- GP

=> y² = xz -- equation (3)

${x}^{b - c} \times {y}^{c - a} \times {z}^{a - b} = 1 \\ \\ Substitute \: \:(b - c)\: \: value \: here, \\ \\ {x}^{a - b} \times {z}^{a - b} \times {y}^{c - a} = 1$

By using aⁿ * bⁿ = (ab)ⁿ,

${(xz)}^{(a - b)} \times {(y)}^{(c - a)} = 1 \\ \\ Substitute \: \:(xz)\: \:value \: here, \\ \\ { ({y}^{2}) }^{(a - b)} \times {y}^{(c - a)} = 1 \\ \\ {(y)}^{(2a - 2b)} \times {y}^{(c - a)} = 1 \: \: \: ( { {a}^{m} })^{n} = {a}^{mn}$

Now,

By using a^m* a^n = a^ (m + n)

${y}^{(2a - 2b + c - a)} = 1 \\ \\ {y}^{(a +c - 2b )} = 1 \\ \\ Substitute \:\: (a + c) \:\: value \: here,\\ \\ {y}^{(2b - 2b)} = 1 \\ \\ {y}^{0} = 1 \\ \\ 1 = 1 \: \: ( {a}^{0} = 1)$

Hence, LHS = RHS.