Question

If 4 times of the fourth term of an AP is equal to 11 times the eleven term of it then show that its fifteen term is zero

Answer 1

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

▶ Given :-

→ 4$a_4$ = 11$a_{11}$ .


▶ To prove :-


→ $a_{15}$ = 0 .


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


We have,

=> 4$a_4$ = 11$a_{11}$ .

=> 4( a + 3d ) = 11( a + 10d ) .

=> 4a + 12d = 11a + 110d .

=> 11a - 4a = 12d - 110d .

=> 7a = - 98d .

=> 7a + 98d = 0 .

=> 7( a + 14d ) = 0 .

=> a + 14d = 0/7 .

=> a + ( 15 - 1 )d = 0.

$\huge \boxed{ \boxed{ \blue{ \bf\therefore a_{15} = 0.}}}$



✔✔ Hence, it is proved ✅✅.



THANKS



#BeBrainly.

Answer 2

=> 4(a4) = 11(a11) .

=> 4( a + 3d ) = 11( a + 10d ) .

=> 4a + 12d = 11a + 110d .

=> 11a - 4a = 12d - 110d .

7a = - 98d .

7a + 98d = 0 .

7( a + 14d ) = 0 .

a + 14d = 0/7 .

a + ( 15 - 1 )d = 0.

•°• a15 = 0.