If 7 times the 7^{text{th}} term of an AP is | vedclass

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(B) Given that $7$ times the $7^{	ext{th}}$ term is equal to $11$ times the $11^{	ext{th}}$ term.$7 a_{7} = 11 a_{11}$Using the formula for the $n^{

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(B) Given that $7$ times the $7^{	ext{th}}$ term is equal to $11$ times the $11^{	ext{th}}$ term.$7 a_{7} = 11 a_{11}$Using the formula for the $n^{	ext{th}}$ term of an $AP$,$a_{n} = a + (n - 1)d$:$7[a + (7 - 1)d] = 11[a + (11 - 1)d]$$7(a + 6d) = 11(a + 10d)$$7a + 42d = 11a + 110d$Rearranging the terms to one side:$11a - 7a + 110d - 42d = 0$$4a + 68d = 0$Dividing by $4$:$a + 17d = 0$We need to find the $18^{	ext{th}}$ term $(a_{18})$:$a_{18} = a + (18 - 1)d = a + 17d$Since $a + 17d = 0$,therefore $a_{18} = 0$.

If $7$ times the $7^{\text{th}}$ term of an $AP$ is equal to $11$ times its $11^{\text{th}}$ term,then its $18^{\text{th}}$ term will be

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