Question

find x, if (p-q) x,(p+q) are in AP​

Answer 1

Answer:

The condition in AP is

2b=a+c

Here,a=p-q

b=x

And,c=p+q

2x=p-q+p+q

2x=2p

x=p

Step-by-step explanation:

Hope it helps you.....

Answer 2

Answer:

plz " FOLLOW " me ....

plz " BRAINLIEST " me....

Step-by-step explanation:

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SEE ANSWERSLog in to add commentAnswers UnusualUnusual ExpertAnswer is one . as x y z are in gp so

SEE ANSWERSLog in to add commentAnswers UnusualUnusual ExpertAnswer is one . as x y z are in gp so y = \sqrt{xz } = > y^{2} = xz

SEE ANSWERSLog in to add commentAnswers UnusualUnusual ExpertAnswer is one . as x y z are in gp so y = \sqrt{xz } = > y^{2} = xz............ (1)

SEE ANSWERSLog in to add commentAnswers UnusualUnusual ExpertAnswer is one . as x y z are in gp so y = \sqrt{xz } = > y^{2} = xz............ (1)now p q r are in ap so

SEE ANSWERSLog in to add commentAnswers UnusualUnusual ExpertAnswer is one . as x y z are in gp so y = \sqrt{xz } = > y^{2} = xz............ (1)now p q r are in ap so q-r = -d. r-p = 2d and p-q = -d

SEE ANSWERSLog in to add commentAnswers UnusualUnusual ExpertAnswer is one . as x y z are in gp so y = \sqrt{xz } = > y^{2} = xz............ (1)now p q r are in ap so q-r = -d. r-p = 2d and p-q = -dso.

SEE ANSWERSLog in to add commentAnswers UnusualUnusual ExpertAnswer is one . as x y z are in gp so y = \sqrt{xz } = > y^{2} = xz............ (1)now p q r are in ap so q-r = -d. r-p = 2d and p-q = -dso. x ^{ - d} .y^{2d} . {z}^{ - d} = {x}^{ - d} {z}^{ - d} . {y}^{2d}

SEE ANSWERSLog in to add commentAnswers UnusualUnusual ExpertAnswer is one . as x y z are in gp so y = \sqrt{xz } = > y^{2} = xz............ (1)now p q r are in ap so q-r = -d. r-p = 2d and p-q = -dso. x ^{ - d} .y^{2d} . {z}^{ - d} = {x}^{ - d} {z}^{ - d} . {y}^{2d} so it becomes

SEE ANSWERSLog in to add commentAnswers UnusualUnusual ExpertAnswer is one . as x y z are in gp so y = \sqrt{xz } = > y^{2} = xz............ (1)now p q r are in ap so q-r = -d. r-p = 2d and p-q = -dso. x ^{ - d} .y^{2d} . {z}^{ - d} = {x}^{ - d} {z}^{ - d} . {y}^{2d} so it becomes = {y^2 / xz}^d = 1^d = 1 ...... from eqn (1)

SEE ANSWERSLog in to add commentAnswers UnusualUnusual ExpertAnswer is one . as x y z are in gp so y = \sqrt{xz } = > y^{2} = xz............ (1)now p q r are in ap so q-r = -d. r-p = 2d and p-q = -dso. x ^{ - d} .y^{2d} . {z}^{ - d} = {x}^{ - d} {z}^{ - d} . {y}^{2d} so it becomes = {y^2 / xz}^d = 1^d = 1 ...... from eqn (1)4.1

SEE ANSWERSLog in to add commentAnswers UnusualUnusual ExpertAnswer is one . as x y z are in gp so y = \sqrt{xz } = > y^{2} = xz............ (1)now p q r are in ap so q-r = -d. r-p = 2d and p-q = -dso. x ^{ - d} .y^{2d} . {z}^{ - d} = {x}^{ - d} {z}^{ - d} . {y}^{2d} so it becomes = {y^2 / xz}^d = 1^d = 1 ...... from eqn (1)4.162 votes

SEE ANSWERSLog in to add commentAnswers UnusualUnusual ExpertAnswer is one . as x y z are in gp so y = \sqrt{xz } = > y^{2} = xz............ (1)now p q r are in ap so q-r = -d. r-p = 2d and p-q = -dso. x ^{ - d} .y^{2d} . {z}^{ - d} = {x}^{ - d} {z}^{ - d} . {y}^{2d} so it becomes = {y^2 / xz}^d = 1^d = 1 ...... from eqn (1)4.162 votesIf there is any confusion please leave a comment below.