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Question

The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.

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Solution

We know that

Tn=a+(n1)d

Given, T4=a+(41)d=0

a+3d=0

a=3d

T25=a+(251)d

= a+24d=(3d)+24d

= 21d

And, T11=a+(111)d

= a+10d

Then, 3T11=3(a+10d)

= 3a+30d

= 3(3d)+30d (a=3d)

= 30d9d=21d=T25

3T11=T25 Hence Proved.

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