three numbers whose sum is 21 are in ap. if the product of the first and third term exceeds the second number by 6. find the numbers
Answers
Answered by
71
Let the 3 numbers be (a-d), a and (a + d).
According to the question,
a-d+a+a+d=21
then 3a=21
a=7
Also,
(a-d)(a+d)=a+6
a^2-d^2=a+6
from above a=7,putting here we get
d=+-6
If d = 6, then the numbers are 7 - 6, 7, 7 + 6
= 1, 7, 13
If d = -6, then the numbers are 7 - (-6), 7, 7 + (-6)
= 13, 7, 1
According to the question,
a-d+a+a+d=21
then 3a=21
a=7
Also,
(a-d)(a+d)=a+6
a^2-d^2=a+6
from above a=7,putting here we get
d=+-6
If d = 6, then the numbers are 7 - 6, 7, 7 + 6
= 1, 7, 13
If d = -6, then the numbers are 7 - (-6), 7, 7 + (-6)
= 13, 7, 1
Answered by
55
this is the answer
if it helped u then mark it as BRAINLIEST
if it helped u then mark it as BRAINLIEST
Attachments:
Similar questions