Question

if the 9th term of an AP is 0 , prove that its 29th term is twice its 19th term​

Answer 1

Answer:-

Given:

9th term of an AP = 0

We know that,

nth term of an AP – a(n) = a + (n - 1)d

⟶ a + (9 - 1)d = 0

⟶ a + 8d = 0

⟶ a = - 8d -- equation (1)

Now,

We have to prove that:

a(29) = 2 * a(19)

⟶ a + (29 - 1)d = 2 * [ a + (19 - 1)d ]

⟶ a + 28d = 2a + 36d

Substitute the value of a from equation (1).

⟶ - 8d + 28d = 2( - 8d) + 36d

⟶ 20d = - 16d + 36d

⟶ 20d = 20d

Hence, Proved.

Answer 2

Question:-

➡ If the 9th term of an A.P. is zero then prove that, 29th term is twice the 19th term.

Proof:-

Let us assume that,

➡ First term of the A.P. = a and,

➡ Common Difference = d

Now,

Nth term of an A.P. = a + (n -1)d

So,

9th term = a + (9 - 1)d

= a + 8d

Now, it's given that, 9th term of the A.P. is zero.

➡ a + 8d = 0 .....(i)

Now,

29th term = a + (29 - 1)d

= a + 28d

19th term = a + (19 - 1)d

= a + 18d

Now,

29th term - 2 × 19th term

= a + 28d - 2 × (a + 18d)

= a + 28d - 2a - 36d

= -a - 8d

= -1(a + 8d)

= -1 × 0

= 0

Hence,

29th term - 2 × 19th term = 0

➡ 29th term = 2 × 19th term. (Hence Proved)