Find five numbers in A.P. such that their sum is 25 and the product of the first and last term is 15
Answers
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.
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
Learn more:
Find the sum of frist 51 terms of the AP whose 2nd term is 2 and 4th ...
https://brainly.in/question/7655866
the 9th and 19th terms of a series of a.p are respectively 35 and 75 ...
https://brainly.in/question/7568849