Question

What is the common difference of four terms in an AP such that the ratio of the
product of the first and fourth terms to that of the second and third is 2:3 and the
sum of all four terms is 20?
​

Answer 1

Answer:

• Common difference = ±2

Step-by-step explanation:

Given:

• In an AP the ratio of the product of the first and fourth terms to that of the second and third terms is 2:3.
• Sum of all four terms = 20.

To Find:

• Common difference of four terms in an AP.

Let, four terms of an AP are (a - 3d), (a - d), (a + d) and (a + 3d).

Now, according to the question,

∴ Sum of four terms = 20

⇒ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 20

⇒ a - 3d + a - d + a + d + a + 3d = 20

⇒ 4a = 20

⇒ a = 20/4

⇒ a = 5

Now, it is also given that,

∴ The ratio of the  product of the first and fourth terms to that of the second and third is 2:3.

$\sf{\implies \dfrac{(a-3d)(a+3d)}{(a-d)(a+d)}=\dfrac{2}{3}}$

$\sf{\implies \dfrac{a^{2}+3da-3da-9d^{2}}{a^{2}+ad-da-d^{2}}=\dfrac{2}{3}}$

$\sf{\implies \dfrac{a^{2}-9d^{2}}{a^{2}-d^{2}}=\dfrac{2}{3}}$

$\sf{\implies 3a^{2}-27d^{2}=2a^{2}-2d^{2}}$

$\sf{\implies a^{2}=25d^{2}}$

Now, put the value of 'a' in above equation,

$\sf{\implies a^{2}=25d^{2}}$

$\sf{\implies (5)^{2}=25d^{2}}$

$\sf{\implies 25=25d^{2}}$

$\sf{\implies d^{2}=\dfrac{25}{25}}$

=> d² = 1

=> d = ± 1

So, four terms be:

• (a - 3d) = (5 - 3) = 2
• (a - d) = (5 - 1) = 4
• (a + d) = (5 + 1) = 6
• (a + 3d) = (5 + 3) = 8

Hence, Common difference = ±2

Answer 2

$\sf{\underline{\boxed{\large{\blue{\mathsf{Solution}}}}}}$

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

• ratio of the product of the first and fourth terms to that of the second and third is 2:3
• sum of all four terms is 20

_______________________

$\sf{\bold{\green{\underline{\underline{To\:Find}}}}}$

• Common difference = ????

______________________

$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

let the 4 consecutive terms be ( a - 3d ) , ( a - d ) ,( a + d ) , ( a + 3d )

• sum of the terms is 20

$\sf\implies a - 3d + a - d + a + d + a + 3d = 20$

$\sf\implies 4a = 20$

$\sf\implies a = \dfrac{20}{4}$

$\sf\implies a = 5$

$\sf{\red{\boxed{\bold{\dag a = 5 }}}}$

• ratio of the product of the first and fourth terms to that of the second and third is 2:3

$\sf\implies\dfrac{ ( a - d ) ( a + d )}{ ( a - 3d) ( a + 3d)} = \dfrac{3}{2}$

$\sf{\underline{\underline{\large{\bold{( a + b ) ( a - b ) = a^2- b^2 }}}}}$

$\sf\implies\dfrac{ a^2 - d^2 }{ a^2 - 9d^2} = \dfrac{3}{2}$

$\sf\implies 2a^2 - 2d^2 = 3a^2 - 27d^2$

$\sf\implies 3a^2 + 2a^2 = 27d^2 - 2d^2$

$\sf\implies 5a^2 = 25 d^2$

• putting value of a

$\sf\implies 5^2 = 25d^2$

$\sf\implies 25 = 25d^2$

$\sf\implies d^2 = \dfrac{25}{25}$

$\sf\implies d = \pm 1$

• finding the no.s

$\sf\longrightarrow a - 3d = 5 - 3\times 1= 5 - 3 = 2$

$\sf\longrightarrow a + 3d = 5 + 3\times 1= 5 + 3 = 8$

$\sf\longrightarrow a - d = 5 - 1 = 4$

$\sf\longrightarrow a + d = 5 + 1 = 6$

• Finding common difference

$\sf\dag 4 - 2 = 2$

$\sf\dag 6 - 4 = 2$

$\sf\dag 8 - 6 = 2$

______________________

$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

• Common difference is 2