Sum of three term in ap is 21 the product of first and third term exceeds the second term by 6
Answers
Answer:
Hence, when,
(I) d = + 6 and a = 7
Numbers = 1 , 7 , 13
(II) d = - 6 and a = 7
Numbers = 13 , 7 , 1
Step-by-step explanation:
Given terms are in AP, let the terms be a - d, a and a + d.
According to the question,
( a - d ) + a + ( a + d ) = 21
=> 3a = 21
=> a = 21 / 3
=> a = 7
We are also provided that,
[ ( a - d ) * ( a + d ) ] - a = 6
Substituting the value of "a" as calculated above,
[ ( 7 - d ) * ( 7 + d ) ] - 7 = 6
=> [ 7² - d² ] - 7 = 6
=> 49 - d² = 6 + 7
=> d ² = 49 - 13
=> d = √36
=> d = ±6
=> d = + 6, - 6
Hence, when,
(I) d = + 6 and a = 7
Numbers = 1 , 7 , 13
(II) d = - 6 and a = 7
Numbers = 13 , 7 , 1
Answer:
Suppose the three term in ap = a ,a + d ,a + 2d .
According to question :
= a + a + d + a + 2d = 3a + 3d = 21.
= ie.. a + d = 7 and d = 7 - a .
= a(a + 2d) - (a + d) = 6 .
= a{a + 2(7 - a )} - 7 = 6 .
= { a + 14 - 2a} = 13 .
= a{14 - a } = 13 .
= 14a - a^2 = 13 .
= a^2 - 14a + 13 = 0 .
= Applying quadratic formula :
a = 13,1.
d = 7 - a ie.. -6 , 6 .
Therefore there are two Ap :
13 ,7,1 or 1,7,13 .