The $4^{th}$ term of an A.P. is zero. Prove that the $25^{th}$ term of the A.P. is three times its $11^{th}$ term.
Given: The 4th term of an A.P. is zero.
To do: To Prove that the 25th term of the A.P. is three times its 11th term.
Solution:
4th term of an $A.P.\ =\ a_{4} =0$
$\therefore \ a\ +\ ( 4\ –\ 1) d\ =0$
$\therefore \ a\ +\ 3d\ =\ 0$
$\therefore \ a\ =\ –3d\ \ \ \ \ \ \dotsc .( 1)$
25th term of an A.P. , $a_{25}$
$=a+( 25\ –\ 1) d$
$=–3d\ +\ 24d\ \ \ \ \ \ \ \ \ \ \dotsc .[ From\ ( 1)]$
$=21d$
11th term of the given A.P. $a_{11} .=a+( n-1) d$
$=-3d+( 11-1) d$
$=-3d+10d$
$=7d$
3 times 11th term of an A.P. $=3a_{11} =3\times 7d=21d$
i.e., the $25^{th}$ term of the A.P. is three times its $11^{th}$ term.
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