Find the two numbers in AP whose sum is 20 and the sum of whose square

1.	Find the two numbers in AP whose sum is 20 and the sum of whose squares is 120.
Answer» Let the required numbers are a – 3d, a – d, a + d and a + 3d which are in AP with common difference 2d. Given that sum of these required numbers is 20. ∴ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20 ⇒ 4a = 20 ⇒ a = \(\frac{20}4\) = 5. Hence, a = 5. Also given that the sum of squares of required numbers is 120. ∴ (a − 3d)² + (a − d)² + (a + d)² +(a + 3d)² = 120 ⇒ a² – 6ad + 9d² + a² – 2ad +d ² +a² + 2ad + d² + a² + 6ad + 9d² =120 ⇒ 4a² + 20d² = 120 ⇒ 4 × 5² + 20d² = 120 ⇒ 20d² = 120 – 4 × 25 = 120 – 100 = 20 ⇒ d² = \(\frac{20}{20}\) = 1 ⇒ d = ±1. Case I :- If d = 1 and a = 5. Then the number are a – 3d, a – d, a + d, a + 3d = 5 – 3, 5 – 1, 5 + 1, 5 + 3 = 2, 4, 6, 8 . Case II:- If d = –1 and a = 5. Then the numbers are a – 3d, a – d, a + d, a + 3d = 5 – 3 ×–1, 5 – (–1), 5 – 1, 5 + 3 (–1) = 5 + 3, 5 + 1, 5 – 1, 5 –3 = 8, 6, 4, 2. Hence the required numbers are 2, 4, 6 and 8.

Find the two numbers in AP whose sum is 20 and the sum of whose squares is 120.

Answer»

Let the required numbers are a – 3d, a – d, a + d and a + 3d which are in AP with common difference 2d.

Given that sum of these required numbers is 20.

∴ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20

⇒ 4a = 20 ⇒ a = \(\frac{20}4\) = 5.

Hence, a = 5.

Also given that the sum of squares of required numbers is 120.

∴ (a − 3d)² + (a − d)² + (a + d)² +(a + 3d)² = 120

⇒ a² – 6ad + 9d² + a² – 2ad +d ² +a² + 2ad + d² + a² + 6ad + 9d² =120

⇒ 4a² + 20d² = 120

⇒ 4 × 5² + 20d² = 120 ⇒ 20d² = 120 – 4 × 25 = 120 – 100 = 20

⇒ d² = \(\frac{20}{20}\) = 1

⇒ d = ±1.

Case I :- If d = 1 and a = 5.

Then the number are a – 3d, a – d, a + d, a + 3d

= 5 – 3, 5 – 1, 5 + 1, 5 + 3 = 2, 4, 6, 8 .

Case II:- If d = –1 and a = 5.

Then the numbers are a – 3d, a – d, a + d, a + 3d

= 5 – 3 ×–1, 5 – (–1), 5 – 1, 5 + 3 (–1) = 5 + 3, 5 + 1, 5 – 1, 5 –3 = 8, 6, 4, 2.

Hence the required numbers are 2, 4, 6 and 8.