IIT JEE , AIEEE Forum - Mathematics , Algebra

scadez

can u pls solve this problem for me......

show that for an A.P,a1,a2,.....,a2k,

a1^2-a2^2+a3^2-a4^2+.............+a2k-1^2-a2k^2= k/(2k+1) *(a1^2-a2k^2)

is this a difficult problem??

if not , i cant even solve this....my tution sir gives me problems like this(sometimes more difficult)....does this mean i dont have a chance in iit?

pls reply soon

rates assured

iitkgp_bipin

Let the common difference be d.
a₂ - a₁ = a₃ - a₂ = ........ = a_2k - a_2k-1 = d

a₁² - a₂² = (a₁+a₂)(a₁-a₂) = (a₁+a₂)(-d)
a₃² - a₄² = (a₃+a₄)(a₃-a₄) = (a₃+a₄)(-d)
......
......
a_2k-1² - a_2k² = (a_2k-1+a_2k)(a_2k-1-a_2k) = (a_2k-1+a_2k)(-d)

Now add all these terms and take (-d) common, the sum becomes :

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

Now the series inside the bracket is sum of an AP.
Apply sum = (n/2)(1st term + last term)

S = (-d){(2k/2)(a₁+a_2k)}

We know that (2k)th term is : a_2k = a₁ + (2k-1)d

From this : (-d) = (a₁-a_2k)/(2k-1)

Now substitute this in the expression of S :

S = {(a₁-a_2k)/(2k-1)}.{(2k/2)(a₁+a_2k)}

S = {k/(2k-1)}.(a₁²-a_2k²)

Don't be demoralized. Keep on practicing.

scadez

dat was a gud 1..........

actually now i feel much better......i only dint get dat last part......

can i improve myself by practice???

i wanna get into iit at any cost!!!!!!!!

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