find the four consecutive term of an AP such that the sum of the middle two term is 18 and the product of the two and is 45
Answers
Answer:
3, 7 , 11 ,15
Step-by-step explanation:
Given ---> Four consecutive terms are in
AP and sum of middle terms is 18 and product of end terms is 45
To find---> Four consecutive terms of AP-
Solution--->
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Let four consecutive term of an AP be (a - 3d ) , ( a - d ) , ( a + d ) and ( a + 3 d )
ATQ
Sum of middle terms = 18
p
Middle two terms are (a - d ) and ( a + d )
So
( a - d ) + ( a + d ) = 18
=> 2 a = 18
=> a = 18 / 2
=> a = 9
Now product of two end terms = 45
End terms are ( a - 3 d ) and ( a + 3 d )
So ( a - 3 d ) ( a + 3 d ) = 45
We have an identity
( p + q ) ( p - q ) = p² - q²
Applying it here
( a )² - (3 d )² = 45
a² - 9 d² = 45
Putting a = 9 here we get
=> (9 )² - 9 d² = 45
=> 81 - 9 d² = 45
=> 81 - 45 = 9 d²
=> 9 d² = 36
=> d ² = 36 / 9
=> d² = 4
=> d = ± 2
If d = 2 and a = 9
a - 3 d = 9 - 3 ( 2 )
= 9 - 6 = 3
a - d = 9 - 2 = 7
a + d = 9 + 2 = 11
a + 3 d = 9 + 3 ( 2 )
= 9 + 6 = 15
So terms are 3 ,7 , 11, 15
If d = -2
a - 3 d = 9 - 3 ( -2 )
= 9 + 6 = 15
a - d = 9 - ( -2 ) = 9 + 2 = 11
a + d = 9 + (- 2 )
= 9 - 2 = 7
a + 3 d = 9 +3 (-2)
= 9 - 6 = 3
So terms are 15 ,11 , 7 , 3
Answer:
Step-by-step explanation:
Let 4 terms are a-3d,a-d,a+d,a+3d
Then a-d+a+d=18
2a=18
a=9
(a-3d) (a+3d)=45
(9-3d)(9+3d)=45
9²-(3d)²=45
81-9d²=45
-9d²=45-81
-9d²=(-36)
9d²=36
d²=4
d=2
hence,
(a-3d)= 9-3×2=9-6=3
(a-d)=9-2=7
(a+d)=9+2=11
(a+3d)=9+3×2=9+6=15
therefore terms are 3,7,11,15 (when d positive)
and 15,11,7,3(when d is negative)