Question

Divide 20 into 4 parts which are in A. P. and such that the product of first and fourth is to product of the second and third in the ratio 2 : 3. [Ans. 2,4,6,8 या 8,6,4,2] ​

Answer 1

Answer:

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Divide 20 into four parts which are in A.P. and such that the product of the first and fourth is to the product of the second and third in the ratio 2:3.

Medium

Solution

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Let the four parts be a – 3d, a – d, a + d and a +3d

Hence (a – 3d) + (a – d) + (a + d) + (a +3d) = 20

⇒ 4a = 20

∴ a = 5

It is also given that (a – 3d)(a + 3d) : (a – d)(a + d) = 2 : 3

(a

2

–d

2

)

(a

2

–9d

2

)

=

3

2

3(a

2

–9d

2

)=2(a

2

–d

2

)

⇒ 3a

2

–27d

2

=2a

2

–2d

2

⇒ 3a

2

–2a

2

=27d

2

–2d

2

⇒ a

2

=25d

2

⇒ 25=25d

2

⇒ d

2

=1

∴d=±1

Case (i): If d=1

Hence (a–3d)=(5–3)=2

(a–d)=(5–1)=4

(a+d)=(5+1)=6

(a+3d)=(5+3)=8

Hence the four numbers are 2, 4, 6 and 8.

Case (ii): If d=–1

Hence (a–3d)=(5+3)=8

(a–d)=(5+1)=6

(a+d)=(5–1)=4

(a+3d)=(5–3)=2

Hence the four numbers are 8, 6, 4 and 2.

Answer 2

Answer:

The A.P. is 2,4,6,8 or 8,6,4,2.

Step-by-step explanation:

Given,

4 parts of 20 are in A.P.

Let the four terms of the A.P be $(a-3d),(a-d),(a+d) and (a+3d)$.

→$(a-3d)+(a-d)+(a+d)+(a+3d)=20$

Opening the brackets and rearranging like terms,

→$4a=20$

→$a=5$

It is also given that, the product of first and fourth terms to the product of second and third terms are in the ratio $2:3$.

⇒$(a-3d)(a+3d):(a-d)(a+d)=2:3$

Using the identity $(a+b)(a-b)=a^{2} -b^{2}$,

⇒$(a^{2}-9d^{2} ):(a^{2} -d^{2} )=2:3$

⇒$\frac{(a^{2}-9d^{2} )}{(a^{2} -d^{2} )} =\frac{2}{3}$

Cross-multiplying on both sides,

⇒$3(a^{2} -9d^{2} )=2(a^{2} -d^{2} )$

⇒$3a^{2} -27d^{2} =2a^{2} -2d^{2}$

⇒$a^{2} =25d^{2}$

Substituting the value of $a$,

⇒$25d^{2} =25$

⇒$d^{2} =1$

⇒$d =1$ or $d =-1$

The terms of the A.P are $(a-3d),(a-d),(a+d)$ and $(a+3d)$

Substituting when $a=5$ and $d=1$,

the A.P is $2,4,6,8$.

Substituting when $a=5$ and $d=-1$,

the A.P. is $8,6,4,2$.