Question

3 term and are in AP such that their sum is 12 product is 120​

Answer 1

Step-by-step explanation:

Let the second term be x and common difference be d

A/Q.

(x -d) +x +(x +d) =12

=> 3x =12

=> x = 12÷3

= 4

so, x= 4

Again,

(x -d) × x × ( x + d) =120

$= > (x {}^{2} - d {}^{2)} \times x = 120$

$= > (4 {}^{2} - d {}^{2} ) \times 4 = 120$

$= > 64 - 4d {}^{2 } = 120$