find a if a2+2a+2, 3a2+6a+6, 4a2+5a+4 are in A.P.
dhathri123:
is that square?
yes
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hi friend,
we know that if p,q,r are in AP
then 2q=p+r
similarly,
2(3a²+6a+6)=a²+2a+2+4a²+5a+4
6a²+12a+12=5a²+7a+6
a²+5a+6=0
a²+2a+3a+6=0
a(a+2)+3(a+2)=0
(a+3)(a+2)=0
a=-3,-2
I hope this will help u:)
we know that if p,q,r are in AP
then 2q=p+r
similarly,
2(3a²+6a+6)=a²+2a+2+4a²+5a+4
6a²+12a+12=5a²+7a+6
a²+5a+6=0
a²+2a+3a+6=0
a(a+2)+3(a+2)=0
(a+3)(a+2)=0
a=-3,-2
I hope this will help u:)
oh
so can i connect u here?
yes ofcourse
thank u so much..(≧▽≦)
:)
by the way how do i know tht 3 variables are in A.P? i mean what shoul be the answer if tgey are ib A.P
if the common difference is same then the numbers are said to be in ap
ohh....ok
thanks
:)
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