Question

If ninth term of an AP is zero, prove that its twenty ninth term is double the
nineteenth term.
​

Answer 1

QUESTION:

If ninth term of an AP is zero, prove that its twenty ninth term is double the

nineteenth term.

ANSWER:

We used the formula here;

$nth \: term(Tn) = a + (n - 1)d$

where;

a = first term

d = common difference

GIVEN:

$T9=0$

that is;

$t9 = a + (9 - 1)d \\ 0 = a + 8d \\ a = - 8d$

hence,

$a = - 8d$

now,

we have to prove :

T29 = 2 × T19

Taking LHS :

$t29 = a + (29 - 1)d \\ (putting \: the \: value \: of \: a) \\ t29 = - 8d + 28d \\ t29 = 20d$

taking RHS;

$2 \times t19 \\ 2 \times (a + (19 - 1)d \\ (again \: putting \: the \: value) \\ 2 \times ( - 8d + 18d) \\ 2 \times 10d$

hence proved;

twenty ninth term is double the nineteenth term.

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Answer 2

Answer :

Given -

• a(9) = 0

To Prove -

• a(29) = 2*a(19)

Solution -

Formula : a + (n - 1)d = a(n)

Firstly, a(9) = a + (9 - 1)d

⇒ a + 8d = 0

⇒ a = - 8d ....(1)

Now, a(29) = a + (29 - 1)d

⇒ a(29) = a + 28d

By Equation (1),

⇒ a(29) = - 8d + 28d

⇒ a(29) = 20d

Now, a(19) = a + (19 - 1)d

⇒ a(19) = a + 18d

By Equation (1),

⇒ a(19) = - 8d + 18d

⇒ a(19) = 10d

Now, We know that, 2*10d = 20d.

Hence, 2*a(19) = a(29).

Hence Proved.