For an A.P.,the 6^{th} term is 19 and the 17^ | vedclass

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(B) The $n^{th}$ term of an $A.P.$ is given by the formula $a_n = a + (n - 1)d$,where $a$ is the first term and $d$ is the common difference.Given,the

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$87$

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(B) The $n^{th}$ term of an $A.P.$ is given by the formula $a_n = a + (n - 1)d$,where $a$ is the first term and $d$ is the common difference.Given,the $6^{th}$ term $a_6 = 19$,so $a + 5d = 19$ (Equation $1$).Given,the $17^{th}$ term $a_{17} = 41$,so $a + 16d = 41$ (Equation $2$).Subtracting Equation $1$ from Equation $2$:$(a + 16d) - (a + 5d) = 41 - 19$$11d = 22$$d = 2$Substituting $d = 2$ into Equation $1$:$a + 5(2) = 19$$a + 10 = 19$$a = 9$Now,find the $40^{th}$ term $a_{40}$:$a_{40} = a + (40 - 1)d$$a_{40} = 9 + 39(2)$$a_{40} = 9 + 78$$a_{40} = 87$.

For an $A.P.$,the $6^{th}$ term is $19$ and the $17^{th}$ term is $41$. Find the $40^{th}$ term of the $A.P.$

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