IIT JEE , AIEEE Forum - Mathematics , Algebra - Arithematic progression........................ one really irritatin prob .......

nano0101

Give me ways of solvin it ........... i dont want answers cos i already have them!!! I just want the method of solvin it .. just the logic ...

Show that for an AP a₁, a₂ , a₃ .....................

S = a²₁- a²₂+ a²₃- a²₄+ .............+ a²_2k-1 - a²_2k

= [k/(2k-1)] (a²₁- a²_2k)

AND

The first term of an AP a_1,a_2,a₃,.......... is equal to unity .

At what value of common difference is a₁ a₃+ a₃ a₂is minimum ?

The answer is -5/4 ..

but the question how exactly do u arrive at it ? I tried

but got the answer to be -2 or -1/2 which is wrong!

KAB

Q1)

Show that for an AP a₁, a₂ , a₃ .....................

S = a²₁- a²₂+ a²₃- a²₄+ .............+ a²_2k-1 - a²_2k

= [k/(2k-1)] (a²₁- a²_2k)

Solution:

S=(a₁-a₂)(a₁+a₂)+(a₃-a₄)(a₃+a₄)+................+(a_2k-1-a_2k)(a_2k-1+a_2k)

Let common difference of the series be d.

So d=a₂-a₁=a₃-a₂=.......=a_2k-a_2k-1..................(1)

Also a_2k=a₁+(2k-1)d

So -d=(a₁-a_2k)/(2k-1).................(2)

So S= -d(a₁+a₂+....+a_2k) (From (1))

= -d((2k/2)(a₁+a_2k))

=(k/(2k-1))(a₁²-a_2k²)

(From(2))

nano0101

thanks a lot .........

KAB

Q2)

The first term of an AP a_1,a_2,a₃,.......... is equal to unity .

At what value of common difference is a₁ a₃+ a₃ a₂is minimum ?

Given a₁=1

Let d be the common difference.

Then a₂=1+d and a₃=1+2d

Let S=a₁ a₃+ a₃ a₂

So a₁ a₃+ a₃ a₂=1+2d+1+2d+d+2d²

or S=2d²+5d+2

So S to be minimum dS/dd=0

So 4d+5=0

or d= -5/4

nano0101

Thanks again!!!