Question

X, Y, Z, are in A. P. and x, y, (z+1) are in G. P. Then

Answer 1

Answer:

Step-by-step explanation:

x, y, z, are in ap

Let the common difference =d

Then y - d = x  -------------(1)

and y + d = z  ------------(2)

x, y, (z+1) are in G. P. Then

y/x = (z + 1) / y  

y^2 = x * (z + 1)  

Putting value of x and z from (1) and (2) we get

y² = (y - d) * (y + d + 1)  

y²= y^2 + yd + y - yd - d² - d  

y² = y^2 + y - d^2 - d  

0 = y - d^2 - d  

y = d² + d

y=d(d+1)--------------------------(3)

Now Let's take d=1 then

y=1*2=2

From(1) and (2)

x=y-d

=2-1=1

and z=y+d

2+1=3

Therefore x=1,y=2,z=3

Then One solution

1,2,3

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We also observe

x=y-d

=d² + d-d=d²

and z=y+d

=d² + d+d

=d² + 2d

hence the series becomes:

d²,d²+d,d²+2d.......................

Hence this is an AP with

first term d² and common difference d

hence infinite series are formed like

d=2  the AP 4,6,8,......................

d=3, then AP is 9,12,15...............

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