Question

find five consecutive in AP whose sum is 45 and the sum of the cube of 2nd and 4th term is 1944

Answer 1

"Find five consecutive terms in A.P. whose sum is 45 and the sum of the cubes of the second and fourth terms is 1944.?" 

Let the five terms be: 

a 
a + d 
a + 2d 
a + 3d 
a + 4d 

Where d is the constant between each consecutive term. 

a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 45 

5a + 10d = 45 

Dividing everything by 5: 

a + 2d = 9 

Now subtract 2d from both sides. The reason will become clear in a moment: 

a = 9 - 2d 

We also know that "the sum of the cubes of the second and fourth terms is 1944". The second term is a + d and the fourth term is a + 3d 

(a + d)³ + (a + 3d)³ = 1944 

Substituting 9 - 2d for a, we get: 

((9 - 2d) + d)³ + ((9 - 2d) + 3d)³ = 1944 

(9 - d)³ + (9 + d)³ = 1944 

Now we know that (a + b)³ = a³ + 3a²b + 3ab² + b³, and we can use that to expand both terms: 

(729 + 3(81)(-d) + 3(9)(d²) - d³) + (729 + 3(81)(d) + 3(9)(d²) + d³) = 1944 

(729 - 243d + 27d² - d³) + (729 + 243d + 27d² + d³) = 1944 

Now combining like terms and canceling where appropriate: 

1458 + 54d² = 1944 

Subtracting 1458 from both sides: 

54d² = 486 

d² = 9 

d = 3 

Going back to a = 9 - 2d 

a = 9 - 2(3) = 9 - 6 = 3 

So the progression is 3, 6, 9, 12, and 15 <== ANSWER 

Let's check to see if it works: 

3 + 6 + 9 + 12 + 15 = 45 <== CHECK 

6³ + 12³ = 216 + 1728 = 1944 <== CHECK 

I hope that helps. Good luck!

Answer 2

Answer:

The five consecutive terms are 3, 6, 9, 12, and 15, which satisfy the given AP pattern.

Step-by-step explanation:

Let us assume the five consecutive terms in arithmetic progression (AP) as a - 2d, a - d, a, a + d, and a + 2d. Then, the sum of these terms is given by 5a, and the sum of the cube of the 2nd and 4th term is (a - d)^3 + (a + d)^3.

Using the given conditions, we can form two equations:

5a = 45 ...(1)

(a - d)^3 + (a + d)^3 = 1944 ...(2)

Solving equation (1), we get a = 9. Substituting this value in equation (2), we get:

(-d)^3 + (d)^3 = 216

2d^3 = 216

d^3 = 108

d = 3∛108

Therefore, the five consecutive terms are 3, 6, 9, 12, and 15, which form an AP with a common difference of 3.

Their sum is 45, and the sum of the cube of the 2nd and 4th term is 1944.

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