The 4th term is of an AP is zero .Prove that its 25th term is triple its 11th term
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Solution:-
Let 'a' be the first term and 'd' be the common difference of the given AP.
Given : a₄ = 0
⇒ a + (4 - 1)d = 0
= a + 3d = 0
a = - 3d................(1)
a₂₅ = a + (25 - 1)d = a + 24d
a₁₁ = a + (11 - 1)d = a + 10d
a₂₅/a₁₁ = a+24d/a+10d
Now, substituting the value of a = - 3d in the above, we get.
= - 3d+24d/- 3d+10d
= 21d/7d
= 3
therefore, a₂₅ = 3*a₁₁
Hence proved.
Let 'a' be the first term and 'd' be the common difference of the given AP.
Given : a₄ = 0
⇒ a + (4 - 1)d = 0
= a + 3d = 0
a = - 3d................(1)
a₂₅ = a + (25 - 1)d = a + 24d
a₁₁ = a + (11 - 1)d = a + 10d
a₂₅/a₁₁ = a+24d/a+10d
Now, substituting the value of a = - 3d in the above, we get.
= - 3d+24d/- 3d+10d
= 21d/7d
= 3
therefore, a₂₅ = 3*a₁₁
Hence proved.
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