The sum of the first three numbers in an ap is 18. If the product of the first and the third term is 5 times the common difference, find the three numbers.
Answers
Answer:
When a = 6 and d = 4 , A.P. : 2 , 6 , 10
When a = 6 and d = -9 , A.P. : 15 , 6 , -3
Step-by-step explanation:
Let the three numbers in an A.P. are a - d, a , a + d
First term = a
Common difference = d
Now, sum of three numbers is 18
⇒ a - d + a + a + d = 18
⇒ 3a = 18
⇒ a = 6
Therefore, The firs term is 6
Now, product of the first and the third term is 5 times the common difference
⇒ (a - d)·(a + d) = 5d
⇒ a² - d² = 5d
⇒ d² + 5d - 36 = 0
⇒ d² + 9d - 4d - 36 = 0
⇒ (d - 4)·(d + 9) = 0
⇒ d = 4 or d= -9
Hence, the possible cases of three numbers are :
When a = 6 and d = 4 , A.P. : 2 , 6 , 10
When a = 6 and d = -9 , A.P. : 15 , 6 , -3
If a = 6 and d = –9, then the first three numbers are 15, 6, –3.
If a = 6 and d = 4, then the first three numbers are 2, 6, 10.
Step-by-step explanation:
Let the first three numbers of an AP are (a – d), a and (a + d).
Sum of the first three numbers = 18
a – d + a + a + d = 18
3a = 18
a = 6
Product of first and third term = 5 times common difference
(a – d)(a + d) = 5d
a² – d² = 5d
6² – d² = 5d
36 – d² = 5d
⇒ d² + 5d – 36 = 0
(d + 9)(d – 4) = 0
d + 9 = 0, d – 4 = 0
d = –9, d = 4
If a = 6, d = –9, then
a – d = 6 – (–9) = 15
a + d = 6 + (–9) = –3
If a = 6 and d = –9, then the first three numbers are 15, 6, –3.
If a = 6, d = 4, then
a – d = 6 – 4 = 2
a + d = 6 + 4 = 10
If a = 6 and d = 4, then the first three numbers are 2, 6, 10.
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