Question

22. If (b - c)2, (c-a)2, (a - b)2 are in A. P. then show that
a re in A. P.​

Answer 1

$\huge\underline\mathfrak\red{Answer}$

${(c - a)}^{2} - {(b - a)}^{2} = {(a - b)}^{2} - {(c - a)}^{2}....(1) \\ \\ if \: \: {(b - c)}^{2} {(c - a)}^{2} and \: {(a - b)}^{2} \: are \: in \: ap \\ \\ solve \: (1) \\ \\ 2 {(c - a)}^{2} = {(a - b)}^{2} + {(b - a)}^{2} \\ \\ take \: negative \: sign \: common\\ \\ 2 {(c - a)}^{2} = {(a - b)}^{2} + {( - b + a)}^{2} \\ \\ 2 {(c - a)}^{2} = {(a - b)}^{2} + {(a - b)}^{2} \\ \\ 2 {(c - a)}^{2} = 2 {(a - b)}^{2} \\ \\ {(c - a)}^{2} = {(a - b)}^{2} \\ \\ (c - a) = (a - b) \\ \\ c + b = 2a \\ \\ so \: acc \: to \: this\: equation \: c \: a \: b \: are \: in \: ap$

Answer 2

i don't know this answer hope before answer will help u lot..

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