the 4th term of an AP is 0.prove that the 25th term of the AP is three times its 11th term
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a4=0
a+3d=0
Now,
a25=a+24d
=>a25=a+3d+21d
=>a25=0+21d
=>a25=21d
a11=a+10d
=>a11=a+3d+7d
=>a11=0+7d
=>a11=7d
=>3(a11)=21d
So,a25=3(a11)
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