there are four terms in Ap.the sum of second and third term is 22 and the product of first and fourth term is 85 . Determine the term In AP
Answers
When the consecutive terms of series differ by a common number, then the series is said to be Arithmetic Progression
Let a be the first term of the AP
d be the common difference of the AP
nth term of AP ⇒ a + ( n - 1) d
Given,
The sum of second and third term is 22
⇒ (a + d) + (a + 2d) =22
⇒ 2a + 3d = 22
⇒ d = 1/3 ( 22 - 2a)
The product of first and fourth term is 85
⇒ a ( a + 3d) = 85
⇒ a² + 3ad = 85
Substituting the value of d gives,
⇒ a ( a + 3 ( 1/3 * (22 - 2a) )) = 85
⇒ a ( a + 22 - 2a) = 85
⇒ a ( - a + 22) = 85
⇒ - a² + 22a = 85
⇒ a² - 22a + 85 = 0
⇒ a² - 17a - 5a + 85 = 0
⇒ a ( a - 17) - 5 ( a - 17)= 0
⇒ (a-5)(a-17)= 0
⇒ a = 5 or a = 17
If a = 5,
d = 1/3 ( 22 - 10) = 1/3 ( 12) = 4
If a = 17,
d = 1/3 ( 22 - 34) = 1/3 ( - 12) = - 4
a = 5, d = 4
Then Arithmetic Progression is
5, 9, 13, 17
a = 17, d = - 4
Then Arithmetic Progression is,
17, 13, 9, 5
Therefore, The required terms in the AP are 5, 9, 13, 17.
Answer:
AP1 : 5, 9, 13, 17
AP2 : 17, 13, 9, 5
Step-by-step explanation:
AP = a1 , a2 , a3 , a4, a5
AP = a , a+d , a+2d , a+3d , a+4d
the sum of 2nd term and 3rd term is 22
a2 + a3 = a+d + a+2d = 22
2a + 3d = 22
3d = 22 - 2a
3d = 2(11 - a)
d = 2/3 (11 - a) ------ {i}
the product of 1st term and 4th term is 85
a1 × a4 = a(a + 3d) = 85
a² + 3ad = 85 ------ {ii}
from {i} and {ii}
a² + 3a × 2/3 (11 - a) = 85
a² + 2a(11 - a) = 85
a² + 22a - 2a² = 85
22a - a² = 85
a² - 22a + 85 = 0
a² - 5a - 17a +85 = 0
a(a - 5) - 17(a - 5) = 0
(a - 17) (a - 5) = 0
a = 5 ,17
If a = 5,
d = 1/3 ( 22 - 10) = 1/3 ( 12) = 4
If a = 17,
d = 1/3 ( 22 - 34) = 1/3 ( - 12) = - 4
If a = 5, d = 4
Then Arithmetic Progression is
5, 9, 13, 17
If a = 17, d = - 4
Then Arithmetic Progression is,
17, 13, 9, 5
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