Question

Find five numbers in A.P. such that their sum is 25 and the product of the first and last term is 15

Answer 1

AnsWer :

• 5 - √10 , 5 - √5/2 , 5 , 5 + √5/2 , 5 + √10.

SolutioN :

Let,

• First term " a "
• Common difference " d "

A/Q,

Case 1.

• Five numbers in A.P. such that their sum is 25
• [ a - 2d , a - d , a , a + d , a +2d. ]

→ a - 2d +a - d +a + a + d + a +2d = 25.

→ 5a = 25.

→ a = 25/5

→ a = 5.

Case 2.

• The product of the first and last term is 15
• [ a - 2d , a + 2d ]

→( a - 2d )( a + 2d ) = 15.

→ a² - 4d² = 15.

→ 5² - 4d² = 15.

→ 25 - 4d² = 15.

→ - 4d² = 15 - 25.

→ - 4d² = - 10.

→ 4d² = 10.

→ 2d² = 5.

→ d² = 5/2.

→ d = ±√5/2.

★ Our AP Become,

# Case 1st, When a = 5. and d = √5/2.

• a - 2d , a - d , a , a + d , a + 2d
• 5 - √10 , 5 - √5/2 , 5 , 5 + √5/2 , 5 + √10.

# Case 2nd, When a = 5. and d = - √5/2.

• a - 2d , a - d , a , a + d , a + 2d.
• 5 + √10 , 5 + √5/2 , 5 , 5 - √5/2 , 5 - √10.

Therefore, the first five number of A.P is 5 - √10 , 5 - √5/2 , 5 , 5 + √5/2 , 5 + √10.

Answer 2

Given : Five numbers in A.P. such that their sum is 25 and the product of the first and last term is 15

To find :  5 numbers in AP

Solution:

Let say  numbers  in AP are

5 - 2d  , 5 - d  , 5  , 5  + d  , 5 + 2d

so that  Sum is 25

now ( 5 - 2d)(5 + 2d) = 15

=> 25 - 4d² = 15

=> 4d² = 10

=> d² = 5/2

=> d = ± √5/√2

Hence five numbers are

5  - √10    ,   5 -  √5/2   , 5  ,  5 + √5/2   , 5  + √10

5  - √10    ,   5 -  √5/2   , 5  ,  5 + √5/2   , 5  + √10

are 5 numbers in AP whose sum = 25 and product of first & last term = 15

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