Question

Divide 69 into three parts which are in A.P and are such that the product of the 1st two parts is 483.

Answer 1

Solution:-

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  

Answer.

Answer 2

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Let the three parts be a-d, a, a+d.

Now,

a-d+a+a+d=69

3a=69

a=23

Product of the first two parts=483

=>(a-d)(a)=483

=>a²-ad=483

=>(23)²-(23)(d)=483

=>529-23d=483

=> -23d=483-529

-23d= -46

d=2

a=23, d=2

So, the AP will be:

a-d=23-2=21

a=23

a+d=25

Therefore, 69=21+23+25.