Question

If 19th term of an AP is zero. Prove that 29th term is doubled of its 19th term

Answer 1

a19=0
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 2

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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