Question

the sum of three numbers in arithmetic progression is 45 and the sum of the squares of the smallest and largest of these numbers is 468. find the largest number.

Answer 1

➡️Solution:

For such situations it is convenient to take three terms as

a-d,a,a+d

According to the question,sum of these terms are 45,

So

$a - d + a + a + d = 45 \\ \\ 3a = 45 \\ \\ a = \frac{45}{3} \\ \\ a = 15$
Now,sum of the squares of the smallest and largest of these numbers is 468.So

$( {a - d)}^{2} + ( {a + d)}^{2} = 468 \\ \\ {a}^{2} + {d}^{2} - 2ad + {a}^{2} + {d}^{2} + 2ad = 468 \\ \\ 2 {a}^{2} + 2 {d}^{2} = 468 \\ \\ {a}^{2} + {d}^{2} = 234 \\ \\ put \: a = 15 \\ \\ {d}^{2} = 234 - 225 \\ \\ {d}^{2} = 9 \\ \\ d = ±3$

So, the numbers can be

$a - d = 15 - 3 = 12 \\ a = 15 \\ a + d = 15 + 3 = 18$
So, those numbers are 12,15,18

Hope it helps you

Answer 2

Answer:

18.

Step-by-step explanation:

We know the general form of an arithmetic progression is (a), (a + d), (a + 2d), (a + 3d) and so on.

Where a is the first term and d is the common difference.

Acc to the question,

(a) + (a+d) + (a+2d) = 45.

or, 3a + 3d = 45.

or, a + d = 15.

or, a = 15 - d.

Now, sum of the squares of the smallest and largest of these numbers is 468.

i.e. (a)² + (a+2d)² = 468

or, (15 - d)² + (15 - d + 2d)² = 468.

or, 225 - 30d + d² + 225 + 30d + d² = 468.

or 450 + 2d² = 468.

or, 2d² = 18

or, d² = 9

or, d = 3.

So, a will be 15 - 3 = 12.

Therefore, the largest term will be (12+2*3) = 18. [Ans]