Question

If a ; b ; c are in A.P.; then b-a =
(a) a+c/2 (b) c-b (c) b^2-1 (d) a-b

Answer 1

Answer:

$(a)..(b)..(c).....are \: in \: ap \\ therefore... = common \: difference \: will \: be \: equal \\ d2 - d1 = d3 - d2 \\ b - a = c - b \\ \\ answer \: is \: (b) = (c - b)$

Answer 2

QUESTION:

If a ; b ; c are in A.P.; then b-a =

(a) a+c/2 (b) c-b (c) b^2-1 (d) a-b

ANSWER:

$a ,\: \: b, \: \: c$

are in AP so,

$b - a = c - b$

Your answer is option b.

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