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The sum of the three numbers in an arithmetic progression is 45 and the sum of the squares of the smallest and largest numbers of these is 468. Find the largest number among these numbers.
1. 15
2. 16
3. 18
4. 21

1 Answer

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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.

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