Three numbers, a, b, and c, none of them zero, form an arithmetic progression.?

Then

2b = a + c ......[Apply concept of arithmatic mean]

squaring

4b^2 = (a+c)^2

a+1,b,c are in GP

b^2 = (a+1)*c ......[Apply concept of geometric mean]

a,b,c+2 are in GP

b^2 = a(c+2) ......[Apply concept of geometric mean]

Thus (a+1)*c = a(c+2) = [(a+c)^2]/4

ac+c = ac+2a

c=2a

4(a+1)c = (a+c)^2

Put c=2a

4(a+1)2a = (a+2a)^2

8a^2 + 8a = 9a^2

a^2 - 8a = 0

a(a-8) = 0

since a is non zero

_____

a=8

_____

c=2*8 = 16

_____

b= (8+16)/2 = 12

_____

Thus

b=12

_____

Answer check:

8,12,16 in A.P.

9,12,16 in G.P.

8,12,18 in G.P.

The numbers a,b,c fofm an arithmetic progression with sum a+b+c=341 a-1,b+2,c+13 form a geometric progression. Find the sum of the members of the geometric progression

First three: x, x+d, x+2d, where x is the first number, d is the difference Last three: x+d, (x+d)r, (x+d)r^2, where r is the ratio. x + x + 2d = 2 => x+d = 1 x+d + (x+d)r^2 = 26 => 1 + r^2 = 26 => r = 5 Since x+2d = (x+d)r => 1+d = r = 5 => d = 4 x+d = 1 =>x+4 = 1 => x = -3 Therefore, the four numbers are: -3,1,5,25

Arithmetic progression: b-a=c-b, then 2b=a+c

Geometric progression: b/(a+1)=c/b, b/a=(c+2)/b

or b^2=(a+1)c=(c+2)a or b^2=ca+c=ca+2a, then 2a=c

Combining, 2b=a+c=a+2a=3a, b=1,5a, so

b/(a+1)=c/b, 1,5a/(a+1)=2a/1,5a, (1,5a)^2=(a+1)2a or

(9/4)a^2 -2a^2-2a=0, (a^2)/4-2a=0, a^2-8a=0, a=8, b=12, c=16

since a,b,c are in AP

therefore, b-a=c-b

this gives us, b=(a+c)/2

if a is increased by 1, we have GP,

thus,

b= sq. root (ac+c)

and when c is increased by 2, we will have

b= sq. root(ac+2a)

we now have 3 eqns. in 3 variables

solving them we get

a= 8

b=12

c=16

so ur answer would be

b=12

1]

a,b,c are in A.P

therefore

b-a=c-b

a+c=2b........1

2]

a+1,b,c are in G.P

therefore

b/[a+1]=c/b

b^2=c[a+1]......2

3]

a,b,c+2 are in G.P

b/a=[c+2]/b

b^2=a[c+2].........3

Now we solve these for a,b,c

from 2&3

c[a+1]=[c+2]a

ac+c=ac+2a

c=2a.......4

from 4&1

2b=a+c

2b=3a

b=3a/2......5

substituting these values in 3

[3a/2]^2=[2a+2]a

9a^2/4=2a^2+2a

9a/4=2a+2

9a=6a+8

a=8

b=12

c=16

verify

8,12,16 are in AP ok

9,12,16 are in GP ok

8,12,18 are in GP ok

course they can if 2b=a+c

4 ur 2nd question- a, b, 2c r in GP,

then, bsquare= a 2c

bsquare= 2ac

therfore, b= sqare root of 2ac

hope i got the ans right :-]

yes if

b-a=c-b