Question

If 8th term of an AP is twice the 13th term then show that second term is twice the 10th term

Answer 1

Answer:

Given :

If tₙ is nth term, then,

t₈ = 2t₁₃

To prove :

t₂ = 2t₁₀

Solution :

We know that

$\boxed{\boxed{\begin{array} {cc} \mathbf{T_n = a + (n-1)d}\\\text {where}\\\mathrm{T_n = nth\ term}\\\mathrm{a = first\ term}\\\mathrm{n = number\ of\ terms}\\\mathrm{d = common\ diferrence}\end{array}}}$

So, applying the given formula for given conditions

$\implies \mathrm{t_8 = a + (8-1)d}$

$\implies \mathrm{t_8 = a + 7d}$

and

$\implies \mathrm{t_{13} = a + (13-1)d}$

$\implies \mathrm{t_{13} = a + 12d}$

Now,

t₈ = 2t₁₃

$\implies \mathrm{a + 7d = 2(a + 12d)}$

$\implies \mathrm{a + 7d = 2a + 24d}$

$\implies \mathrm{a + 17d = 0}$ ___[1]

Now, to prove t₂ = 2t₁₀,

Let us consider t₂ = 2t₁₀

then,

$\implies \mathrm{t_2 = a + (2-1)d}$

$\implies \mathrm{t_2 = a + d}$

and

$\implies \mathrm{t_{10} = a + (10-1)d}$

$\implies \mathrm{t_{10} = a + 9d}$

Now,

t₂ = 2t₁₀

$\implies \mathrm{a + d = 2(a + 9d)}$

$\implies \mathrm{a + d = 2a + 18d}$

$\implies \mathrm{a + 17d = 0}$

This consideration satisfies the given condition [1], So t₂ = 2t₁₀ is proved

Hence proved!!

Hope it helps!!