Question

Prove that 21st term of a.p is twice it's 11th term if first term is 0

Answer 1

Given:

The 1st term of an AP is 0, i.e., a = 0.

Prove that:

The 21st term is twice the 11th term.

Solution:

$n^{th} \; term \; of \; an \; AP \; is :\\\\a + (n-1) d$

Since both the terms are of the same AP, the common difference d will remain the same.

11th term:

a + (n-1)d.

0 + (11-1)d.

10d.

21st term:

a + (n-1)d.

0 + (21-1)d.

20d.

t₂₁ = 2t₁₁.

20d = 2(10d).

20d =20d.

∴ It is proved that the 21st term of an Arithmetic Progression is 2 times the 11th term of an arithmetic Progression if it's 1st term is 0.

Answer 2

Given that the first term of the AP is zero

To prove

$\sf{{a}_{21}=2{a}_{11}}$

As the first term is zero,the AP would be:

a,a + d,a + 2d,...........

→0,d,2d,...........

Thus,the common difference is d

Now,

Nth term: a + (n-1)d

When n = 11,

a + (11-1)d

= a + 10d

= 10d.............[1]

When n = 21,

a + (21-1)d

= a + 20d

= 20d..............[2]

Dividing equations [1] and [2],we get the ratio 1:2

Thus, $\sf{{a}_{21}=2{a}_{11}}$

Henceforth,proved