Math, asked by dhruvvaviya7, 17 days ago

Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers.​

Answers

Answered by sandhuraja021
1

Step 1 : Assume the number and follow given instruction

Let the there numbers be a,b,c

As they are in continued proportion

b^ 2=ac

Given that mean proportional is 12

∴b=12

So, 12 ^2 =ac

144=ac

c= a/144

The sum of the remaining two numbers is 26

a+c=26

a+ a/144

=26

Step 2 : Simplify using factorization

On solving we get,

⇒a ^2 +144−26a=0

⇒a ^2−18a−8a+144=0

⇒a(a−18)−8(a−18)=0

⇒(a−18)(a−8)=0

∴a = 18 or a = 8

c = 8 or c = 18

Hence , The numbers are 8,12,18 or 18,12,8

Answered by itzmecutejennei
1

Answer:

Step 1 : Assume the number and follow given instruction

Let the there numbers be a,b,c

As they are in continued proportion

b

2

=ac

Given that mean proportional is 12

∴b=12

So, 12

2

=ac

144=ac

c=

a

144

The sum of the remaining two numbers is 26

a+c=26

a+

a

144

=26

Step 2 : Simplify using factorization

On solving we get,

⇒a

2

+144−26a=0

⇒a

2

−18a−8a+144=0

⇒a(a−18)−8(a−18)=0

⇒(a−18)(a−8)=0

∴a = 18 or a = 8

c = 8 or c = 18

Hence , The numbers are 8,12,18 or 18,12

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