If twice the 11^{th} term of an A.P. is equal | vedclass

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(C) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n-1)d$.$11^{th}$ t

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$0$

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(C) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n-1)d$.$11^{th}$ term of $A.P. = a + 10d$.$21^{st}$ term of $A.P. = a + 20d$.According to the problem,$2 	imes (11^{th} 	ext{ term}) = 7 	imes (21^{st} 	ext{ term})$.$2(a + 10d) = 7(a + 20d)$.$2a + 20d = 7a + 140d$.$5a + 120d = 0$.Dividing by $5$,we get $a + 24d = 0$.The $25^{th}$ term of the $A.P.$ is $a_{25} = a + (25-1)d = a + 24d$.Since $a + 24d = 0$,the $25^{th}$ term is $0$.

If twice the $11^{th}$ term of an $A.P.$ is equal to $7$ times of its $21^{st}$ term,then its $25^{th}$ term is equal to

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