Prove that the product of 2nd and 3rd terms of an AP exceeds the prod

1.	Prove that the product of 2nd and 3rd terms of an AP exceeds the product of 1st and 4th termsby twice the square of the difference between the 1st and 2nd terms.
Answer» Let the A.P. be a, a + d, a + 2d, ..... Here, we have to prove: T2T3- T1T4= 2(T1- T2)^2 Proof: We know that, Tn= a + (n - 1)d LHS = T2T3- T1T4 = (a + d)(a + 2d) - a(a + 3d) = a^2+ ad + 2ad + 2d^2- a^2- 3ad = 2d^2 Now, RHS = 2(T1- T2)^2 = 2(a - (a + d))^2 = 2 (a - a - d)^2 = 2d^2 Hence, LHS = RHS

Prove that the product of 2nd and 3rd terms of an AP exceeds the product of 1st and 4th termsby twice the square of the difference between the 1st and 2nd terms.

Answer»

Let the A.P. be a, a + d, a + 2d, .....

Here, we have to prove: T2T3- T1T4= 2(T1- T2)^2

Proof:

We know that, Tn= a + (n - 1)d

LHS = T2T3- T1T4

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

= a^2+ ad + 2ad + 2d^2- a^2- 3ad

= 2d^2

Now, RHS = 2(T1- T2)^2

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

= 2 (a - a - d)^2

= 2d^2

Hence, LHS = RHS