Question

if nineth term of an A.P is zero. prove that the 29th term is double the 19th term​

Answer 1

Answer:

Let the first term of an A.P. = a

And common difference = d

Given: 9th term of an A.P. is 0. Therefore,

We have to prove that

Thus,

And

From equation (2) and (3),

Hence proved

Answer 2

Answer

• a₂₉ = 2 (a₁₉)

Given

9th term of an AP is 0

To Prove

$\bullet \;\; \bf a_{29}=2(a_{19})$

Proof

nth term of an AP is ,

$\bullet \;\; \bf a_n=a+(n-1)d$

A/c , " ninth term of an A.P is zero "

⇒ a₉ = 9

⇒ a + (9-1)d = 0

⇒ a + 8d = 0

⇒ a = - 8d ... (1)

Let's calculate 19th and 29th term ,

$\rm a_{19}\\\\ \implies \rm a+(19-1)d\\\\\implies \rm a+18d\\\\\implies \rm (-8d)+18d\ \; [From\ (1)]\\\\\implies \rm 10d...(2)$

$\rm a_{29}\\\\\implies \rm a+(29-1)d\\\\\implies \rm a+28d\\\\\implies \rm (-8d)+28d\\\\\rm \implies 20d\\\\\implies \rm 2(10d)\\\\\implies \rm 2(a_{19})$

So ,

$\bf a_{29}=2(a_{19})$

Hence proved