If 9th term of an A.P. is zero, prove that its 29th term is double the 19th term.

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#### Solution

Given:

\[a_9 = 0 \]

\[ \Rightarrow a + \left( 9 - 1 \right)d = 0 \left[ a_n = a + \left( n - 1 \right)d \right]\]

\[ \Rightarrow a + 8d = 0\]

\[ \Rightarrow a = - 8d . . . (i)\]

To prove:

\[a_{29} = 2 a_{19} \]

Proof:

\[\text { LHS }: a_{29} = a + \left( 29 - 1 \right)d\]

\[ = a + 28d\]

\[ = - 8d + 28d \left( \text { From }(i) \right)\]

\[ = 20d\]

\[RHS: 2 a_{19} = 2\left[ a + \left( 19 - 1 \right)d \right]\]

\[ = 2(a + 18d)\]

\[ = 2a + 36d\]

\[ = 2( - 8d) + 36d \left( \text { From }(i) \right)\]

\[ = - 16d + 36d\]

\[ = 20d\]

LHS = RHS Hence, proved.

Concept: Arithmetic Progression (A.P.)

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