Question

find the four terms of an AP whose sum is 16 and the sum whose squares is 84

Answer 1

Answer:

if we let the first term be a and the common difference be d, the four terms are  

a, a + d, a + 2d, a + 3d  

Their sum is 16, so we have  

4a + 6d = 16  

Or  

2a + 3d = 8  

We also have  

a² + (a + d)² + (a + 2d)² + (a + 3d)² = 84  

Since the sum of their squares is 84  

Expanding, we get  

a² + a² + 2ad + d² + a² + 4ad + 4d² + a² + 6ad + 9d² = 84  

Combine like terms  

4a² + 12ad + 14d² = 84  

But since 2a + 3d = 8, squaring both sides we get  

4a² + 12ad + 9d² = 64  

Subtract this from  

4a² + 12ad + 14d² = 84 to get  

5d² = 20  

Divide by 5 and take the square-root to get  

d = ±2  

Plug this into 2a + 3d = 8 to get that either  

2a + 6 = 8 so a = 1  

Or 2a - 6 = 8, so a = 7  

So the two possible arithmetic progressions are  

1, 3, 5, 7  

and  

7, 5, 3, 1, which are actually the same solution.  

Answer 2

Step-by-step explanation:

if we let the first term be a and the common difference be d, the four terms are

a, a + d, a + 2d, a + 3d

Their sum is 16, so we have

4a + 6d = 16

Or

2a + 3d = 8

We also have

a² + (a + d)² + (a + 2d)² + (a + 3d)² = 84

Since the sum of their squares is 84

Expanding, we get

a² + a² + 2ad + d² + a² + 4ad + 4d² + a² + 6ad + 9d² = 84

Combine like terms

4a² + 12ad + 14d² = 84

But since 2a + 3d = 8, squaring both sides we get

4a² + 12ad + 9d² = 64

Subtract this from

4a² + 12ad + 14d² = 84 to get

5d² = 20

Divide by 5 and take the square-root to get

d = ±2

Plug this into 2a + 3d = 8 to get that either

2a + 6 = 8 so a = 1

Or 2a - 6 = 8, so a = 7

So the two possible arithmetic progressions are

1, 3, 5, 7

and

7, 5, 3, 1, which are actually the same solution.