If 7 times the 7^{th} term of an A.P. is equa | vedclass

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

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$0$

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(C) Let $a$ be the first term and $d$ be the common difference of an $A.P.$The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n-1)d$.Therefore,$a_7 = a + 6d$ and $a_{11} = a + 10d$.According to the problem,$7a_7 = 11a_{11}$.Substituting the values,we get $7(a + 6d) = 11(a + 10d)$.Expanding both sides: $7a + 42d = 11a + 110d$.Rearranging the terms: $7a - 11a = 110d - 42d$.$-4a = 68d$,which simplifies to $a = -17d$ (Equation $1$).We need to find the $18^{th}$ term,$a_{18} = a + 17d$.Substituting $a = -17d$ from Equation $1$ into the expression for $a_{18}$:$a_{18} = -17d + 17d = 0$.Thus,the $18^{th}$ term of the $A.P.$ is $0$.

If $7$ times the $7^{th}$ term of an $A.P.$ is equal to $11$ times its $11^{th}$ term,then the $18^{th}$ term of the $A.P.$ is

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