An AP consists of 50 terms of which the 3^{rd | vedclass

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(B) Given that the $3^{rd}$ term $a_3 = 12$ and the total number of terms $n = 50$,with the last term $a_{50} = 106$.Using the formula for the $n^{th}

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$64$

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(B) Given that the $3^{rd}$ term $a_3 = 12$ and the total number of terms $n = 50$,with the last term $a_{50} = 106$.Using the formula for the $n^{th}$ term of an $AP$,$a_n = a + (n - 1)d$:For the $3^{rd}$ term: $a + 2d = 12$ $(i)$For the $50^{th}$ term: $a + 49d = 106$ $(ii)$Subtracting equation $(i)$ from $(ii)$:$(a + 49d) - (a + 2d) = 106 - 12$$47d = 94$$d = 2$Substituting $d = 2$ in equation $(i)$:$a + 2(2) = 12$$a + 4 = 12$$a = 8$Now,find the $29^{th}$ term $a_{29}$:$a_{29} = a + (29 - 1)d$$a_{29} = 8 + 28(2)$$a_{29} = 8 + 56 = 64$Thus,the $29^{th}$ term is $64$.

An $AP$ consists of $50$ terms of which the $3^{rd}$ term is $12$ and the last term is $106$. Find the $29^{th}$ term.

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