Question

The sum of the three numbers in

a.p is 21 and the product of the first and third number of the sequence is 45.

Answer 1

Terms are (a-d), a, (a+d)

Now,

a-d+a+a+d =21

3a =21

a=21/3

a=7

Now,

Product of first and last term

(a+d) * (a-d) =45
---------------------
By formula, a²-b² =(a+b) (a-b)
---------------------------

a² -d² =45

Putting the value of a

7² -d² =45

49-d² =45

49-45 =d²

4=d²

√4 =d

2 =d


Now,


Terms are

First term = (a-d) =7-2 =5

Second term =7

Third term =(a+d) =7+2=9

Terms are :- 5,7,9


I hope this will help you



-by ABHAY

Answer 2

Let the A.P be

a - d , a , a + d

given the sum of A.P is 21

a - d + a + a + d = 21

3a = 21

a = 7
________________

Product of the first and third no. is 45

( a - d ) ( a + d ) = 45

[ using identity ( x - y ) ( x + y ) = x² - y² ]

a² - d² = 45

( 7 )² - d² = 45

49 - d² = 45

- d² = - 4

d = √4

d = ±2

________

The required AP are :-

__________

If a = 7 and d = 2

7 - 2 , 7 , 7 + 2

5 , 7 , 9

_________

If a = 7 and d = - 2

7 + 2 , 7 , 7 - 2

9 , 7 , 5