Question

the sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7:15 . find the numbers​

Answer 1

AnswEr :

Let the terms of AP be (a + 3d), (a + d), (a - d), (a - 3d) where a is the first term and d is the common Difference.

• First Part of the Question Now :

⇒ Sum of Four Terms of AP = 32

⇒ (a + 3d) + (a + d) + (a - d) + (a - 3d) = 32

⇒ 4a + 4d - 4d = 32

⇒ 4a = 32

• Dividing Both term by 4

⇒ a = 8

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• Second Part of the Question Now :

⇒ (1st × 4th) term : (2nd × 3rd) term = 7 : 15

⇒ (a + 3d) × (a - 3d) : (a + d) × (a - d) = 7 : 15

• (a + b)(a - b) = (a² - b²)

⇒ a² - 9d² : a² - d² = 7 : 15

• Product of Extreme = Product of Mean

⇒ 15 × (a² - 9d²) = 7 × (a² - d²)

⇒ 15a² - 135d² = 7a² - 7d²

⇒ 15a² - 7a² = 135d² - 7d²

⇒ 8a² = 128d²

• Dividing Both term by 8

⇒ a² = 16d²

• putting the value of a = 8

⇒ ( 8 )² = 16d²

⇒ 64 = 16d²

• Dividing Both term by 16

⇒ 4 = d²

⇒ d = √4

⇒ d = ± 2

◗ when d is 2 then AP will be ;

↠ (a + 3d), (a + d), (a - d), (a - 3d)

↠ (8 + 3(2)), (8 + 2), (8 - 2), (8 - 3(2))

↠ 14, 10, 6, 2

◗ when d is - 2 then AP will be ;

↠ (a + 3d), (a + d), (a - d), (a - 3d)

↠ (8 + 3(- 2)), (8 + (- 2)), (8 - (- 2)), (8 - 3(- 2))

↠ 2, 6, 10, 14

∴ AP can be both as Shown Above.

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• Tips to Solve Question like this :

• If the number of terms are odd, let the terms be (a-2d), (a-d), a, (a+d), (a+2d) and so on.

• If the number of terms are even, let the terms be (a-3d), (a-d), (a+d), (a+3d) and so on.

Answer 2

$\bf{\Huge{\underline{\boxed{\rm{\red{ANSWER\::}}}}}}$

Given:

The sum of four consecutive numbers in an A.P. is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7:15.

To find:

The numbers.

$\bf{\Large{\underline{Explanation\::}}}}$

Let the four consecutive numbers in A.P. be a-3d,a-d,a+d & a+3d

• First term= a-3d
• Second term= a-d
• Third term= a+d
• Fourth term= a+3d

A/q,

→ a-3d + a-d + a+d + a+3d = 32

→ $a\cancel{-3d}+a\cancel{-d}+a+\cancel{d}+a+\cancel{3d}=32$

→ 4a = 32

→ a = $\cancel{\frac{32}{4} }$

→ a = 8

Now,

The ratio of the product of the first & last term to the product of two middle terms is 7:15.

⇒ $\frac{(a-3d)(a+3d)}{(a-d)(a+d)} =\frac{7}{15}$

⇒ $\frac{a^{2} - 9d^{2}} {a^{2} - d^{2} }=\frac{7}{15}$

⇒ 15(a² - 9d²) = 7(a² - d²)

⇒ 15a² - 135d² = 7a² - 7d²

⇒ 15a² - 7a² = -7d² + 135d²

⇒ 8a² = 128d²

[Putting the value of a in above, we get;]

⇒ 8(8)² = 128d²

⇒ 8 × 64 = 128d²

⇒ 512 = 128d²

⇒ d² = $\cancel{\frac{512}{128} }$

⇒ d² = 4

⇒ d = √4

⇒ d = 2

Now,

• The four consecutive numbers are;

First number= a-3d

⇒ 8 - 3×2

⇒ 8 - 6

⇒ 2

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Second number= a-d

⇒ 8 - 2

⇒ 6

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Third number= a+d

⇒ 8 + 2

⇒ 10

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Fourth number= a+3d

⇒ 8 + 3(2)

⇒ 8 + 6

⇒ 14