Question

. If 4 times the t₄ of an AP is equal to 9 times the t₉, then 13 times the t₁₃ of this AP is
13 times the t₇
13 times the t₆
7 times the t₄
0​

Answer 1

Given that, the t₄ of an AP is 4 times equal to 9 times the t₉ of the AP, 4( t₄ ) = 9( t₉ ).

⌑ We've to find out the 13th term of the AP is 13 times equal to. So, let's Solve —

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As we know that,

The nth formula of an AP is Given by : ${\pmb{\sf{a + (n - 1) d}}}.$

Therefore,

$:\implies\sf a_4 = a + 3d\qquad\bigg\lgroup\sf eq^{n}\; (1)\bigg\rgroup$

Also,

$:\implies\sf a_9 = a + 8d\qquad\bigg\lgroup\sf eq^{n}\;(2)\bigg\rgroup$

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$\underline{\bigstar\:\boldsymbol{According\;to\;the\; Question\::}}$⠀⠀

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• It is given that, the fourth term of the AP is 4 times equal to 9 times the ninth term of the AP.

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✇ Now, from eqₙ ( 1 ) and eqₙ ( 2 ) —

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$\dashrightarrow\sf 4(a + 3d) = 9(a + 8d)\\\\\\\dashrightarrow\sf 4a + 12d = 9a + 72d\\\\\\\dashrightarrow\sf 4a - 9a + 12d - 72d=0\\\\\\\dashrightarrow\sf 5a -60d = 0\\\\\\\dashrightarrow\sf -5(a + 12d) = 0\\\\\\\dashrightarrow\underline{\boxed{\pmb{\frak{\purple{a + 12d = 0}}}}}\;\bigstar$

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$\therefore\:{\underline{\sf{Therefore,\;the\;13\;times\;of\;13^{th}\; term\:of\:AP\:is\:{\pmb{0}}.}}}$

Answer 2

Given :-

If 4 times the t₄ of an AP is equal to 9 times the t₉

To Find :-

then 13 times the t₁₃ of this AP is

Solution :-

We know that

$\sf a_n = a+(n-1)d$

For the 4th term

$\sf a_{4} = a + (4-1)d$

$\sf a_4 = a+3d...(1)$

For the 9th term

$\sf a_{9} = a + (9-1)d$

$\sf a_9 = a+8d...(2)$

Now, ATQ

$\sf 4(a_4) = 9(a_9)$

$\sf 4(a+3d)=9(a+8d)$

$\sf 4a+12d = 9a+72d$

$\sf 4a-9a=72d-12d$

$\sf -5a=60d$

$\sf 5a-60d=0$

$\sf 5(a-12d)=0$

$\sf a-12d=\dfrac{0}{5}$

$\sf a-12d=0$

Now

$\sf a_{13} = a+(13-1)d$

$\sf a_{13}=a+12d$

Hence,

13th term is 0