Question

Sum of three terms in an A.P.is 15 and sum of the squares of the extreme terms is 58. Find the numbers

Answer 1

Answer:

The answer is 3,5 and 7.

Step-by-step explanation:

Let ( a - d ) , a , ( a + d ) are three numbers in A.P

For first condition:

1) The sum of three number is = 15

a - d + a + a + d = 15 (-d and +d cancel out)

3a = 15

a = 15/3

a = 5

Now for 2nd condition:

sum of squares of the extreme terms = 58

Apply the whole square formula.

( a - d )² + ( a + d )² = 58

a² + d² - 2ad + a² +  d² +2ad (after cancellation taking 2 common)

2( a² + d² ) = 58

( a² + d² ) = 58/2

a² + d² = 29

5² + d² = 29  ( a = 5 )

d² = 29 - 25

d² = 4

d = ± 2

Therefore, Required numbers are  

If a = 5 , d = 2

1) a - d = 5 - 2 = 3

a = 5

2) a + d = 5 + 2 = 7

Thus the numbers are 3, 5 and 7.