Where r is common ratio, then what should be value of K to satisfy the equation

Asked by vishakhachandan026 | 23rd Jan, 2020, 04:42: PM

Expert Answer:

Where r is common ratio, then what should be value of K to satisfy the equation
begin mathsize 16px style open vertical bar table row cell straight x to the power of straight k end cell cell straight x to the power of straight k plus 1 end exponent end cell cell straight x to the power of straight k plus 2 end exponent end cell row cell straight y to the power of straight k end cell cell straight y to the power of straight k plus 1 end exponent end cell cell straight y to the power of straight k plus 2 end exponent end cell row cell straight z to the power of straight k end cell cell straight z to the power of straight k plus 1 end exponent end cell cell straight z to the power of straight k plus 2 end exponent end cell end table close vertical bar end style
begin mathsize 16px style straight x to the power of straight k straight y to the power of straight k straight z to the power of straight k open vertical bar table row 1 straight x cell straight x squared end cell row 1 straight y cell straight y squared end cell row 1 straight z cell straight z squared end cell end table close vertical bar equals straight x to the power of straight k straight y to the power of straight k straight z to the power of straight k open vertical bar table row 1 ar cell straight a squared straight r squared end cell row 1 cell ar squared end cell cell straight a squared straight r to the power of 4 end cell row 1 cell ar cubed end cell cell straight a squared straight r to the power of 6 end cell end table close vertical bar end style
begin mathsize 16px style straight a to the power of 3 straight k end exponent straight r to the power of 6 straight k end exponent straight a cubed straight r cubed open vertical bar table row 1 1 1 row 1 straight r cell straight r squared end cell row 1 cell straight r squared end cell cell straight r to the power of 4 end cell end table close vertical bar equals straight a to the power of 3 open parentheses straight k plus 1 close parentheses end exponent straight r to the power of 6 straight k plus 3 end exponent open parentheses 1 minus straight r close parentheses open parentheses straight r minus straight r squared close parentheses open parentheses straight r squared minus 1 close parentheses end style
If k = -1 then we will have Δ = (r - 1)2 (1 - 1/r2)

Answered by Sneha shidid | 28th Jan, 2020, 09:52: AM

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