If 19th term of an AP is zero. Prove that 29th term is doubled of its 19th term
Answers
a+18d=0
a=-18d
a29=2(a19)
a+28d=2(a+18d)
-18d+28d=2(-18d+18d)
10d=2(0)
10d=0
d=0
hence proved
Answer:
HEY MATE HERE'S YOUR ANSWER
Step-by-step explanation:
By given the ninth term of the A.P is 0.
a₉ = 0
The nth term of an A.P is given by,
aₙ = a₁ + (n - 1) × d
where a₁ is the first term,
aₙ is the nth term
d is the common difference
Therefore,
a₉ = a₁ + 8d = 0
a₁ = -8d----(1)
Now the 29th term of the A.P is given by,
a₂₉ = a₁ + 28 d
Substitute the value of a₁ from equation 1,
a₂₉ = -8d + 28 d
a₂₉ = 20 d
a₂₉ = 2 × (10 d)
a₂₉/2 = 10 d----(2)
Now the 19th term of the A.P is given by,
a₁₉ = a₁ + 18 d
Substituting value of a₁ from equation 1,
a₁₉ = -8d + 18 d
a₁₉ = 10d----(3)
From equations 2 and 3, RHS are equal, therefore LHS must also be equal.
Hence,
a₁₉ = a₂₉/2
a₂₉ = 2 × a₁₉
That is the 29th term is double the 19th term.
Hence proved.
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