Correct Answer - Option 3 : 18
Given:
The sum of the three numbers in a series is 45.
The sum of the squares of the smallest and largest number of the series is 468.
Calculation:
Let three terms of an AP series be (a - d), a, and (a + d) where “a” and “d” are the first term and the common difference of an AP.
According to the question,
Sum of three numbers in the series = 45
(a - d) + a + (a + d) = 45
3a = 45
⇒ a = 15 ----(1)
Also, it is given that the sum of the smallest and the largest numbers is 468.
(a - d)² + (a + d)² = 468
⇒ a² + d² - 2ad + a² + d² + 2ad = 468
⇒ 2a² + 2d² = 468 ----(2)
Putting the value of “a” from equation (1) in equation (2),
⇒ 2(15)² + 2d² = 468
⇒ 2 × 225 + 2d² = 468
⇒ 450 + 2d² = 468
⇒ 2d² = 468 – 450 = 18
⇒ d = 3
⇒ a – d = 15 – 3 = 12
⇒ a + d = 15 + 3 = 18
The three terms of the AP series “a - d”, “a” and “a + d” are 12, 15, and 18 respectively.
∴ The largest number among these numbers is 18.