Question

Q. No 6
Find three consecutive terms in an A. P. Whose sum is 27 and their product is 512.​

Answer 1

Answer:

Let the numbers be (a−d),a,(a+d). Then,

Sum=−3⇒(a−d)+a+(a+d)=−3⇒3a=−3⇒a=−1

Product of their cubes=512⇒(a−d)3×a3×(a+d)3=512

⇒(a2−d2)3×a3=512         

⇒−(1−d2)3=512            (∵a=−1)

⇒−(1−d2)=8            

⇒d2=9⇒d=±3    

If d=3, then the numbers are −4,−1,2. If d=−3, then the numbers are 2,−1,−4.

Answer 2

QUESTION:

• Find three consecutive terms in an A. P. Whose sum is -3 and the product of their cubes is 512.

ANSWER:

Given:

• Sum of 3 consecutive terms of an AP = -3
• Product of cubes of 3 consecutive terms of an AP = 512

To Find:

• The 3 consecutive terms

Solution:

Let the 3 consecutive terms be (a - d), (a) and (a + d) respectively.

We are given that,

⇒ Sum of the terms = -3

⇒ (a - d) + (a) + (a + d) = -3

⇒ a - d + a + a + d = -3

Solving the like terms,

⇒ 3a = -3

⇒ a = -1 --------(1)

We are also given that,

⇒ Product of cube of the terms = 512

⇒ (a - d)³ × (a)³ × (a + d)³ = 512

⇒ a³(a - d)³ ×(a + d)³ = 512

We know that,

→ (p+q)(p-q) = p² - q²

So,

⇒ a³(a² - d²)³ = 512

From (1),

⇒ -1³ (-1² - d²)³ = 512

⇒ -1(1 - d²)³ = 512

⇒ (d² - 1)³ = 512

⇒ (d² - 1)³ = 8×8×8

⇒ (d² - 1)³ = 8³

Cube-rooting both sides,

⇒ d² - 1 = 8

⇒ d² = 9

Square-rooting both sides,

⇒ d = ±3

So, terms are:

For d = 3,

• a - d = -1 - 3= -4
• a = -1
• a + d = -1 + 3 = 2

For d = -3,

• a - d = -1 + 3 = 2
• a = -1
• a + d = -1 - 3 = -4

So, the terms are (-4, -1, 2) or (2, -1, -4).

Formula used:

• (a+b)(a-b) = a^2 - b^2