Question

Find four consecutive terms in a.p. whose sum is 72 and ratio of product of the extreme to the product of means is 9:10​

Answer 1

ANSWER:

Given:

• 4 consecutive terms of AP
• Sum of the terms = 72
• Ratio of Product of extremes to product of means = 9:10

To Find:

• The terms.

Solution:

Let the 4 consecutive terms be: a - 3d, a - d, a + d, a + 3d.

We are given that,

⇒ Sum of the terms = 72

Putting the terms,

⇒ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 72

⇒ a - 3d + a - d + a + d + a + 3d = 72

Rearranging the terms,

⇒ a + a + a + a - 3d - d + d + 3d = 72

All the ds get cancelled. So,

⇒ 4a = 72

Transposing 4 to RHS,

⇒ a = 72/4

⇒ a = 18

We are also given that,

⇒ Ratio of Product of extremes to product of means = 9:10

⇒ Product of extremes : Product of means = 9:10

⇒ (a - 3d)(a + 3d) : (a - d)(a + d) = 9:10

We know that,

⇒ (a + b)(a - b) = a² - b²

So,

⇒ [a² - (3d)²] : [a² - d²] = 9:10

⇒ (a² - 9d²) : (a² - d²) = 9:10

⇒ (a² - 9d²)/(a² - d²) = 9/10

On cross multiplying,

⇒ 10(a² - 9d²) = 9(a² - d²)

⇒ 10a² - 90d² = 9a² - 9d²

⇒ 10a² - 9a² = - 9d² + 90d²

⇒ a² = 81d²

⇒ a = √(81d²)

⇒ a = ±9d

So,

⇒ 18 = ±9d

⇒ d = ±2.

Hence,

The terms are (d = 2)

• a - 3d = 18 - 3(2) = 18 - 6 = 12
• a - d = 18 - 2 = 16
• a + d = 18 + 2 = 20
• a + 3d = 18 + 3(2) = 18 + 6 = 24

The terms are (d = -2)

• a - 3d = 18 - 3(-2) = 18 + 6 = 24
• a - d = 18 - (-2) = 18 + 2 = 20
• a + d = 18 + (-2) = 18 - 2 = 16
• a + 3d = 18 + 3(-2) = 18 - 6 = 12

Therefore, the terms are 12, 16, 20 and 24 or 24, 20, 16 and 12.

Verification:

1) Sum is 72.

⇒ 12 + 16 + 20 + 24 ⇒ 28 + 44 ⇒ 72 --------(i)

2) Product of extremes : Product of means = 9:10

⇒ (12×24) : (16×20) ⇒ 288 : 320 ⇒ 9 : 10 --------(ii)

From (i) & (ii),

HENCE VERIFIED!!!

Answer 2

Given:

• Sum of four consecutive terms of A.P. is 72.
• Ratio of Product of the extreme to that of its means is 9:10

To Find:

• The four consecutive terms of the A.P.

Solution:

Let the four consecutive terms of the A.P. be:

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

where,

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

Now, According to the Question,

↦ (a-3d)+(a-d)+(a+d)+(a+3d) = 72

↦ 4a = 72

↦ a = 72/4

↦ a = 18

We know that,

• Product of extremes = (a-3d)(a+3d)
• Product of means = (a-d)(a+d)

Now, According to the Question,

↦ Product of Extreme/Product of Means = 9/10

↦ (a-3d)(a+3d)/(a-d)(a+d) = 9/10

We know that,

• (a+b)(a-b) = a²-b²

By Applying Identity on numerator and denominator in LHS:

↦ a²-(3d)²/a²-d²= 9/10

↦ a²-9d²/a²-d²= 9/10

On Cross Multiplication,

↦ 10(a²-9d²) = 9(a²-d²)

↦ 10a²-90d² = 9a²-9d²

↦ 10a²-9a²-90d² = -9d²

↦ a²-90d² = -9d²

↦ a² = -9d²+90d²

↦ a² = 81d²

↦ a = √81d²

↦ a = ±9d

↦ d = ±a/9

↦ d = ±18/9

↦ d = ±2

Now let us substitute the value of a = 18 and d = 2 in A.P.

• First Term = a-3d = 18-3×2 = 18-6 = 12

• Second Term = a-d = 18-2 = 16

• Third Term = a+d = 18+2 = 20

• Fourth Term = a+3d = 18+3(2) = 18+6 = 24

Now let us substitute the value of a = 18 and d = -2 in A.P.

• First Term = a-3d = 18-3×(-2) = 18+6 = 24

• Second Term = a-d = 18+2 = 20

• Third Term = a+d = 18-2 = 16

• Fourth Term = a+3d = 18+3(-2) = 18-6 = 12