Question

Divide 56 in four parts in AP such that the ratio of the product of their extremes (1st and 4th) to the product of means (2nd and 3rd) is 5:6. ​

Answer 1

Let the four parts be a−3d,a−d,a+d,a+3d

Sum, 4a=56

a=14

${ \: \: \: \: \mapsto\sf \dfrac{(a-3d)(a+3d)}{(a-d)(a+d)}= \dfrac{5}{6} }$

• Using Identity (A+B)(A-B) = A²-B²

${ \: \: \: \: \mapsto\sf \dfrac{ {a}^{2} - {(3d)}^{2} }{ {a}^{2} - {d}^{2} }= \dfrac{5}{6} }$

${ \: \: \: \: \mapsto\sf \dfrac{ {a}^{2} - 9 {d}^{2} }{ {a}^{2} - {d}^{2} }= \dfrac{5}{6} }$

${\: \: \: \: \mapsto\sf 6({a}^{2} - 9 {d}^{2}) = 5( {a}^{2} - {d}^{2}) }$

${\: \: \: \: \mapsto\sf 6{a}^{2} - 54{d}^{2} = 5 {a}^{2} - 5{d}^{2} }$

${\: \: \: \: \mapsto\sf 6{a}^{2} -5 {a}^{2} - 54{d}^{2} + 5{d}^{2}= 0 }$

${\: \: \: \: \mapsto\sf {a}^{2} - 49{d}^{2} = 0 }$

• Putting the value of a

${\: \: \: \: \mapsto\sf {14}^{2} - 49{d}^{2} = 0 }$

${\: \: \: \: \mapsto\sf 196 - 49{d}^{2} = 0 }$

${\: \: \: \: \mapsto\sf - 49{d}^{2} = - 196 }$

${\: \: \: \: \mapsto\sf {d}^{2} = 196 \div 49 }$

${\: \: \: \: \mapsto\sf {d}^{2} = 4 }$

${\: \: \: \: \mapsto\sf d = \sqrt{4} }$

${\: \: \: \: \mapsto\sf d = \pm2 }$

• Numbers are 8,12,16,20.

Answer 2

Answer:

Let the four parts be a−3d,a−d,a+d,a+3d

Sum, 4a=56

a=14

{ \: \: \: \: \mapsto\sf \dfrac{(a-3d)(a+3d)}{(a-d)(a+d)}= \dfrac{5}{6} }↦

(a−d)(a+d)

(a−3d)(a+3d)

=

6

5

Using Identity (A+B)(A-B) = A²-B²

{ \: \: \: \: \mapsto\sf \dfrac{ {a}^{2} - {(3d)}^{2} }{ {a}^{2} - {d}^{2} }= \dfrac{5}{6} }↦

a

2

−d

2

a

2

−(3d)

2

=

6

5

{ \: \: \: \: \mapsto\sf \dfrac{ {a}^{2} - 9 {d}^{2} }{ {a}^{2} - {d}^{2} }= \dfrac{5}{6} }↦

a

2

−d

2

a

2

−9d

2

=

6

5

{\: \: \: \: \mapsto\sf 6({a}^{2} - 9 {d}^{2}) = 5( {a}^{2} - {d}^{2}) }↦6(a

2

−9d

2

)=5(a

2

−d

2

)

{\: \: \: \: \mapsto\sf 6{a}^{2} - 54{d}^{2} = 5 {a}^{2} - 5{d}^{2} }↦6a

2

−54d

2

=5a

2

−5d

2

{\: \: \: \: \mapsto\sf 6{a}^{2} -5 {a}^{2} - 54{d}^{2} + 5{d}^{2}= 0 }↦6a

2

−5a

2

−54d

2

+5d

2

=0

{\: \: \: \: \mapsto\sf {a}^{2} - 49{d}^{2} = 0 }↦a

2

−49d

2

=0

Putting the value of a

{\: \: \: \: \mapsto\sf {14}^{2} - 49{d}^{2} = 0 }↦14

2

−49d

2

=0

{\: \: \: \: \mapsto\sf 196 - 49{d}^{2} = 0 }↦196−49d

2

=0

{\: \: \: \: \mapsto\sf - 49{d}^{2} = - 196 }↦−49d

2

=−196

{\: \: \: \: \mapsto\sf {d}^{2} = 196 \div 49 }↦d

2

=196÷49

{\: \: \: \: \mapsto\sf {d}^{2} = 4 }↦d

2

=4

{\: \: \: \: \mapsto\sf d = \sqrt{4} }↦d=

4

{\: \: \: \: \mapsto\sf d = \pm2 }↦d=±2

Numbers are 8,12,16,20.