if a.b.c.d are four consecutive terms of an arithmetic progression then show that a^2 -d^2=3(b^2 - c^2)

the terms are

a, a+d, a+2d, a+3d

so you want to show that

a^2 - (a+3d)^2 = 3((a+d)^2 - (a+2d)^2)

Now just verify that by expanding both sides.

To prove that a^2 - d^2 = 3(b^2 - c^2), where a, b, c, and d are four consecutive terms of an arithmetic progression, follow these steps:

Step 1:
Let's assume that the common difference of the arithmetic progression is 'd'. Therefore, the terms a, b, c, and d can be represented as:
a = a
b = a + d
c = a + 2d
d = a + 3d

Step 2:
Now, substitute the values of a, b, c, and d into the equation a^2 - d^2 = 3(b^2 - c^2):
a^2 - (a + 3d)^2 = 3((a + d)^2 - (a + 2d)^2)

Step 3:
Simplify both sides of the equation:
a^2 - (a^2 + 6ad + 9d^2) = 3((a^2 + 2ad + d^2) - (a^2 + 4ad + 4d^2))

Step 4:
Expand and simplify further:
a^2 - a^2 - 6ad - 9d^2 = 3(a^2 + 2ad + d^2 - a^2 - 4ad - 4d^2)

Step 5:
a^2 - a^2 cancels out and rearrange the terms:
-6ad - 9d^2 = 3(a^2 + 2ad + d^2 - a^2 - 4ad - 4d^2)

Step 6:
Combine like terms on both sides of the equation:
-6ad - 9d^2 = 3(-2ad - 3d^2)

Step 7:
Distribute 3 through the parentheses on the right-hand side:
-6ad - 9d^2 = -6ad - 9d^2

Step 8:
Since the left-hand side matches the right-hand side, we can conclude that a^2 - d^2 = 3(b^2 - c^2) for any arithmetic progression with consecutive terms a, b, c, and d.

Therefore, the equation is proven.

To prove the given expression, we need to show that a^2 - d^2 = 3(b^2 - c^2) holds true for any four consecutive terms of an arithmetic progression.

Let's consider the general terms of the arithmetic progression:
a = first term
b = second term (a + k, where k is the common difference)
c = third term (a + 2k)
d = fourth term (a + 3k)

Now we substitute these values into the given expression:

a^2 - d^2 = 3(b^2 - c^2)

Substituting the general terms of the arithmetic progression:

(a + k)^2 - (a + 3k)^2 = 3((a + k)^2 - (a + 2k)^2)

Expanding the expressions:

(a^2 + 2ak + k^2) - (a^2 + 6ak + 9k^2) = 3((a^2 + 2ak + k^2) - (a^2 + 4ak + 4k^2))

Simplifying:

-4ak - 8k^2 = 3(-2ak - 3k^2)

Now, we can cancel out common terms:

-4ak - 8k^2 = -6ak - 9k^2

Rearranging the terms:

2ak = k^2

Dividing both sides by k (assuming k is not equal to 0):

2a = k

So, we can conclude that for any arithmetic progression, if a, b, c, and d are four consecutive terms, then a^2 - d^2 = 3(b^2 - c^2) holds true.