Find four consecutive terms in AP such that the sum of the first and the last is 24 and the product of the second and the third is 135
Answers
Answer:
consecutive terms are 3,9,15,21
Step-by-step explanation:
given --->
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terms are in AP and consecutive and
sum of first and last is 24
product of second and third is 135
To find--->
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four consecutive terms which are in AP
solution --->
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let four consecutive terms in ap are
(a-3d) ,(a-d), (a+d),(a+3d)
now ATQ
sum of first and last term=24
(a-3d) +(a+3d) = 24
-3d and +3d cancel out each other
2a = 24
24
a = ------=12
2
now ATQ
product of second and third term =135
(a-d) (a+d)=135
applying x²-y²=(x+y)(x-y)
(a)²-(d) ² =135
(12)²-d² = 135
144-d²=135
d²=144-135
d²= 9
d =(+3)or(-3)
if d= +3 and a=12
a-3d=12-3(3)=12-9=3
a-d=12-3=9
a+d=12+3=15
a+3d=12+3(3)=12+9=21
so numbers are 3,9,15,21
if d=-3 and a=12
a-3d =12-3(-3)=12+9=21
a- d =12- (-3) =12 + 3=15
a+d =12+(-3)=12-3=9
a+3d =12+3(-3)=12-9=3
so numbers are 21,15,9,3
so in both cases numbers are same
so answer is 3,9,15,21
additional information --->
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1.if there are three consecutive terms of an ap then we can assume them
(a-d),a, (a+d)
2. if there are five consecutive terms in ap then we can assume them
(a-2d), (a-d),a, (a+d),(a+2d)
Step-by-step explanation:
Answer
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