The 4th term of an ap is zero prove that 25th term of an ap is 3 times its 11th term
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1
aₓ=a₁+(x-1)d
a₄=a₁+3d=0
a₂₅=a₁+24d=21d
a₁₁=a₁+10d=7d
a₂₅=3.a₁₁
a₄=a₁+3d=0
a₂₅=a₁+24d=21d
a₁₁=a₁+10d=7d
a₂₅=3.a₁₁
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Given a4= 0
i.e. (a + 3d) = 0
⇒ a = - 3d (1)
nth term of AP = a + (n – 1)d
a11= a + 10d = – 3d + 10d =7d [From (1)]
a25= a+ 24d
= – 3d + 24d
= 21d [From (1)]
= 3 x 7d
Hence a25= 3 x a11
Hope it will help
i.e. (a + 3d) = 0
⇒ a = - 3d (1)
nth term of AP = a + (n – 1)d
a11= a + 10d = – 3d + 10d =7d [From (1)]
a25= a+ 24d
= – 3d + 24d
= 21d [From (1)]
= 3 x 7d
Hence a25= 3 x a11
Hope it will help
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