Question

If the 8th term of an A.P. is zero, then show that 28th term is double the 18th term.

Answer 1

Given:

• a₈ = 0

To Prove:

• a₂₈ = 2a₁₈

Solution:

We know that a term of an AP can be expressed as:

a + (n - 1)d = an ; Where an is the position of the term.

⇒ a₈ = 0

⇒ a + (8 - 1)d = 0

⇒ a + 7d = 0

⇒ a = - 7d

We have to show that;

⇒ a₂₈ = 2a₁₈

⇒ a + 27d = 2[a + 17d]

⇒ -7d + 27d = 2[-7d + 17d]

⇒ 20d = 2[10d]

⇒ 20d = 20d

⇒ LHS = RHS

Hence, proved.

Answer 2

GIVEN :–

• 8th term of A.P. = 0

TO PROVE :–

• 28th term is double the 18th term.

SOLUTION :–

• nth term of A.P. , If first term = a and Common difference = d is –

$\\ \dashrightarrow \large { \boxed { \bold{T_{n} = a + (n - 1)d}}}$

• According to the question –

$\\ \implies { \bold{T_{8} = 0}}$

$\\ \implies { \bold{a + (8 - 1)d = 0}}$

$\\ \implies { \bold{a + 7d = 0}}$

$\\ \implies { \bold{a = - 7d \: \: \: \: - - - eq.(1)}}$

• Now –

$\\ \implies{ \bold{T_{18} = a + (18 - 1)d}}$

$\\ \implies{ \bold{T_{18} = a + 17d}}$

• And –

$\\ \implies{ \bold{T_{28} = a + (28 - 1)d}}$

$\\ \implies{ \bold{T_{18} =a + 27d}}$

• We have to prove –

$\\ \: \: \blacktriangleright \: \: { \bold{T_{28} =2T_{18}}}$

• Let's take L.H.S. –

$\\ \implies { \bold{T_{28} =a + 27d}}$

• Using eq.(1) –

$\\ \implies { \bold{T_{28} = - 7d+ 27d}}$

$\\ \implies { \bold{T_{28} = 20d \: \: \: \: \: - - - eq.(2)}}$

• Let's take R.H.S. –

$\\ \implies { \bold{2T_{18} = 2 \{a + 17d \}}}$

$\\ \implies { \bold{2T_{18} = 2 a + 34d}}$

• Using eq.(1) –

$\\ \implies { \bold{2T_{18} = 2( - 7d) + 34d}}$

$\\ \implies { \bold{2T_{18} = - 14d + 34d}}$

$\\ \implies { \bold{2T_{18} = 20d \: \: \: \: - - - eq.(3)}}$

• By eq.(2) & eq.(3) –

$\\ \dashrightarrow \: \: { \bold{L.H.S. = R.H.S.}}$

$\\ \: \: \dashrightarrow \: \: \: \: { \underbrace{ \bold{Hence \: \: proved}}}$