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If a1,a,a3...an are in A.P then prove th...

If `a_1,a_,a_3...a_n` are in A.P then prove that `a_1^2-a_2^2+a_3^2-a_4^2+....a_(2k-1)^2-a_(2k)^2= (k/(2k-1))(a_1^2-a_(2k)^2)`

A

` k/(2k-1)(a_(1)^(2) - a_(2k)^(2))`

B

` (2k)/(k-1)(a_(2k)^(2)-a_(1)^(2))`

C

` k/(k+1)(a_(1)^(2)-a_(2k)^(2))`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
a
Since `a_(1),a_(2),….,a_(n)` are in AP, therefore
`a_(2)-a_(1)=a_(3)-a_(2)=...=a_(2k)-a_(2k-1)=d`
Now, ` a_(1)^(2)-a_(2)^(2)=(a_(1)-a_(2))(a_(1)+a_(2))=-d(a_(1)+a_(2))`
` a_(3)^(2) - a_(4)^(2)= d (a_(3) +a_(4))`
` a_(2k-1)^(2)-a_(2k)^(2)=d (a_(2k-1)+a_(2k))`
On adding above equations , we get
` S = -d (a_(1)+a_(2)+...+a_(2k))=-d [(2k)/2 (a_(1)+a_(2k))]`
` = - dk(a_(1)+a_(2k))=(-dk)/((a_(1)-a_(2k)))(a_(1)^(2)=a_(2k)^(2))`
`+(a_(2k-2)-a_(2k-1))+(a_(2k-1)-a_(2k))]`
` = k/(2k-1) (a_(1)^(2)- a_(2k)^(2))`
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MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS-SEQUENCES AND SERIES -EXERCISE 3
  1. If a1,a,a3...an are in A.P then prove that a1^2-a2^2+a3^2-a4^2+....a(2...

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