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If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that xb−c × yc−a × za−b = 1 .

If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that xb−c × yc−a × za−b = 1 .
Solution :
Since a, b and c are in A.P,
b - a  = c - b  = d (common difference)
We need to prove,
xb−c × yc−a × za−b = 1 
Let us try to convert the powers in terms of one variable.
2b  = c + a - a + a
2b  = c - a + 2a
2(b - a)  = c - a
2d  = c - a 
If c - b  = d, then b - c = -d
If b - a  = d, then a - b = -d
L.H.S
xb−c × yc−a × za−b  =  x−d × y2d × z−d   ---(1)
y = √xz
By applying the value of y in (1)
 =  x−d × (√xz)2d × z−d  
 =  x−d × (xz)d × z−d  
  =  x−d + d  z-d + d
  =  1

Hence proved.

Find x, y and z, given that the numbers x, 10, y, 24, z are in A.P.

Find x, y and z, given that the numbers x, 10, y, 24, z are in A.P.
Solution :
Since the given sequence is an arithmetic progression,
t2 - t1  = 10 - x  ---(1)
t3 - t2  = y - 10 ---(2)
t4 - t3  = 24 - y ---(3)
t5 - t4  = z - 24 ---(4)
(1)  = (2)
10 - x  = y - 10
x + y  = 20
(2) = (3)
y - 10  = 24 - y
y + y  = 24 + 10
2y  = 34
y  = 17
By applying the value of y in x + y  = 20
x + 17  = 20
x  = 3
(3)  = (4)
24 - y  = z - 24
y + z  = 24 + 24
y + z  = 48
17 + z  = 48
z  = 31

Hence the values of x, y and z are 3, 17, 31 respectively.

Given A = {a,x,y,r,s}, B = {1,3,5,7,-10},verify the commutative property of set union.

Question 5
Given A = {a,x,y,r,s}, B = {1,3,5,7,-10},verify the commutative property of set union.
Solution:
commutative property of set union
A U B = B U A
A U B = {a,x,y,r,s} U {1,3,5,7,-10}
    =  {a,x,y,r,s,1,3,5,7,-10}   ------ (1)
B U A = {1,3,5,7,-10} U {a,x,y,r,s}
    =  {a,x,y,r,s,1,3,5,7,-10}  ------ (2)

(1) = (2)