Question

If a1, a2, a3,...,a2k are in AP, prove , a1²-a2²+a3²-a4²+...+a2k-1²-a2k² = k(a1²- a2k²)/(2k-1)​

Answer 1

Answer:

Since, a

1

,a

2

,a

3

,.....a

2n

form an AP.

Therefore,

a

2

−a

1

=a

4

−a

3

=....a

2n

−a

2n−1

=d

Here, a

1

2

−a

2

2

+a

3

2

−a

4

2

+....+a

2n−1

2

−a

2n

2

=(a

1

−a

2

)(a

1

+a

2

)+(a

3

−a

4

)(a

3

+a

4

)+....+(a

2n−1

−a

2n

)⋅(a

2n−1

+a

2n

)

=−d(a

1

+a

2

+....+a

2n

)=−d(

2

2n

(a

1

+a

2n

)

Also, we know a

2n

=a

1

+(2n−1)d

⇒d=

2n−1

a

2n

−a

1

⇒−d=

2n−1

a

1

−a

2n

∴Therefore, the sum is

=

2n−1

n(a

1

−a

2n

)⋅(a

1

+a

2n

)

=

2n−1

n

⋅(a

1

2

−a

2n

2

).

Step-by-step explanation:

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