Question

if 7 times the 7th term of an AP is equal to 11 times the 11th term, show that the 18th term of it is 0

Answer 1

Answer: $a_{18}$ = 0.

Step-by-step explanation:

$\huge \bf \pink{Hey \: there !! }$

▶ Given :-

→ 7$a_7$ = 11$a_{11}$

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▶ To prove :-

→ $a_{18}$   = 0 .

$\huge \green{ \underline{ \overline{ \bf Solution :- }}}$

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We have,

=> 7$a_7$ = 11$a_{11}$

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=> 7( a + 6d ) = 11( a + 10d ) .

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

=> 11a - 7a = 42d - 110d .

=> 4a = - 68d .

=> 4a + 68d = 0 .

=> 4( a + 17d ) = 0 .

=> a + 17d = 0/4 .

=> a + ( 18 - 1 )d = 0.

$\huge \boxed{ \boxed{ \blue{ \bf\therefore a_{18} = 0.}}}$

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✔✔ Hence, it is proved ✅✅.

THANKS

#BeBrainly.

Answer 2

here your answere ,