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

Question

Increasing a by 1 OR increasing c by 2 results in an geometric progression. Find the value of b.

WOW...a-b=c-b=common difference....I would've never thunk it given this is defined as an A.P.!!!!

This is not my homework, I am still helping my brother with a packet of 100 questions. This is number 76...

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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 :-]

Answer 8

yes if
b-a=c-b