Question

If the 9th term of an A.P. is zero then show that the 29th term is twice the 19th term.​

Answer 1

$\huge {\red{\boxed{ \overline{ \underline{ \mid\mathfrak{An}{\mathrm{sw}{ \sf{er}} \colon\mid}}}}}}$

Given :

• a9 = 0

Solution :

We have formula :

$\large \star {\boxed{\sf{a_n \: = \: a \: + \: (n \: - \: 1)d}}} \\ \\ \implies {\sf{a_{9} \: = \: a \: + \: (9 \: - \: 1)d}} \\ \\ \implies {\sf{0 \: = \: a \: = \: 8d}} \\ \\ \implies {\sf{a \: = \: -8d}}----(1)$

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Use same formula 29 to term :

Substitute value of a from (1)

$\implies {\sf{a_{29} \: = \: a \: + \: (29 \: - \: 1)d}} \\ \\ \implies {\sf{a_{29} \: = \: -8d \: + \: 28d}} \\ \\ \implies {\sf{a_{29} \: = 20d}}$

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Use same formula for 19 to term

Substitute value of a from (1)

$\implies {\sf{a_{19} \: = \: a \: + \: (19 \: - \: 1)d}} \\ \\ \implies {\sf{a_{19} \: = \: -8d \: + \: 18d}} \\ \\ \implies {\sf{a_{19} \: = \: 10d}}$

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So, 29 th term is double of 19th term

Answer 2

Solution :

Let a and d be the first term and common difference of the AP

Since,

$\sf a_9 = 0 \\ \\ \leadsto \sf a + 8D = 0 \\ \\ \leadsto \boxed{\boxed{\sf{a = - 8D}}}$

$\rule{300}{2}$

$\sf a_{29} = a + 28D \\ \\ \leadsto \sf a_{29} = - 8D + 28D \\ \\ \leadsto \sf a_{29} = 20D$

$\rule{300}{2}$

$\sf a_{19} = a + 18D \\ \\ \leadsto \sf a_{19} = - 8D + 18D \\ \\ \leadsto \sf a_{19} = 10 D$

Since,

$\sf a_{29} = 2a_{19}$

Hence,ProveD