Question

Split $69$ late three pats such that they are in $A.P.$ and the product of two smaller parts is $483$.

• A. $21$
• B. $23$
• C. $25$
• D. $Allofabove$

Answer 1

The correct option is D $Allofabove$

Let the first term of the AP be 'a'

And the common difference be 'd'

Since 69 split into 3 parts such that they form an AP.

Let the three parts be (a - d), (a) and (a + d).

Therefore,

(a - d) + (a) + (a + d) = 69

3a = 69

a = 23

The product if two smaller parts = 483

So,

(a) × (a - d) = 483

23 × (23 - d) = 483

⇒ 529 - 23d = 483

⇒ - 23d = 483 - 529

⇒ - 23 d = - 46

⇒ d = 46/23

⇒ d = 2

Therefore,

The 3 parts are

23 - 2 = 21 ;

23

and 23 + 2 = 25

Hence the parts of the given AP are 21, 23, 25