Find four consecutive terms in an AP such that the sum of the second and third terms is 18 and the product of the first and the fourth terms is 45.
Answers
Let the numbers be a - 3 d , a - d , a + d , a + 3 d .
The sum of the second and the third term is 18 .
⇒ a - d + a + d = 18
⇒ 2 a = 18
⇒ a = 18/2
⇒ a = 9
Product of 1st and 4 th term is 45 .
( a + 3 d )( a - 3 d ) = 45
⇒ a² - 9 d² = 45
⇒ 9² - 9 d² = 45
⇒ 81 - 9 d² = 45
⇒ - 9 d² = 45 - 81
⇒ - 9 d² = - 36
⇒ d² = 4
⇒ d = ± 2
When d = + 2 ,
A.P = 9 - 6 , 9 - 2 , 9 + 2 , 9 + 6 .
= 3 , 7 , 11 , 15 .
When d = - 2 .
A.P = 15 , 11 , 7 , 3
Hence the numbers are 3 , 7 , 11 and 15 .
Answer: The four consecutive terms in an AP are 3, 7, 11 and 15 when the sum of the second and third terms is 18 and the product of the first and the fourth terms is 45.
Step-by-step explanation:
Let the four consecutive terms in an AP be a-3d , a-d, a+d , a+3d
According to question the sum of the second and the third term is 18 .
⇒ (a - d )+ (a + d )= 18
⇒ 2a = 18
⇒ a = 18/2
⇒ a = 9
Product of first and fourth term in AP series is 45 .
(a + 3d )( a - 3d ) = 45
⇒ a² - 9d² = 45
⇒ 9² - 9d² = 45
⇒ 81 - 9d² = 45
⇒ - 9d² = 45 - 81
⇒ - 9d² = - 36
⇒ d² = 4
⇒ d = ± 2
Putting d = 2 ,
Putting the values of a and d in A.P series a-3d , a-d, a+d , a+3d = 9 - 6 , 9 - 2 , 9 + 2 , 9 + 6 .
Solving we get 3, 7, 11, 15 .
Putting d = -2 .
Putting the values of a and d in A.P series a-3d , a-d, a+d , a+3d we get = 15 , 11 , 7 , 3
Hence, the four consecutive terms in an AP are 3, 7, 11 and 15 .
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