Question

if 7 times the 7th term of an ap is equal to 11 times the 11th term then what will be its 18th term?​

Answer 1

$\huge\orange{\boxed{\bold{Solution}}}$

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Let a be the first term and D be the common difference of the given AP. Then,

$\bold {7t_{7}} \: = \bold {11t_{11}} \\ \\ \bold {= > 7(a + 6d)} = \bold {11(a + 10d)} \\ \\ \bold {= > 7a + 42d} = \bold {11a + 110d} \\ \\ \bold {= > 4a + (18 - 1)d = 0} \\ \\ \: \bold {= > t_{18}}$

Hence, the 18th term of the given AP is zero.

$\huge\underline\mathfrak\green{Hence\:proved}$

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$\huge\mathbb\blue{THANK\:YUH!}$

Answer 2

$\:\:\:\:\:\:\huge\underline\mathfrak{Solution:}$

We know that,

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

ATQ,

if 7 times the 7th term of an ap is equal to 11 times the 11th term.

.•. 7 × (a+6d) = 11 × (a+10d)

=> 7a + 42d = 11a + 110d

=> -4a = 68d

=> a = -17d .....(1)

Now, 18th term of AP :

$a_1_8$ = a + 17d

$a_1_8$ = -17d + 17d {from (1)}

$a_1_8$ = 0

Therefore,18th term is 0.