Question

The 9th term of an AP is zero. Prove that the 29th term is doubled the 19th term.

Answer 1

$\huge\bigstar\underline\mathfrak\red{Solution}$

Given : The 9th term of an AP is zero.

To prove : $a_2_9$ = $2a_1_9$

Proof : a + 8d = 0 ( given that 9th term is zero ).

where, a is first term of AP and d is common difference.

=> $\huge\boxed{a\:=\:-\:8\:d\:}$

Now, $a_2_9$ = a + 28d ...(i)

put a = -8d in eq (i).

$a_2_9$ = ( - 8d ) + 28d

= -8d + 28d

= 20d

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Also,

$a_1_9$ = a + 18d ....(ii)

put a = -8d in eq (ii)

$a_1_9$ = ( - 8d ) + 18d

= -8d + 18d

= 10d

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( $a_2_9$ )/ ( $a_1_9$ )

= 20d / 10d

= 2 / 1

By cross multiplying it,

We get,

$a_2_9$ = $2a_1_9$

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Hence proved!

Answer 2

SOLUTION:-

Refer to the attachment.

Hope it helps ☺️