If the 9^{th} and 19^{th} terms of an arithme | vedclass

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(B) Let the first term be $a$ and the common difference be $d$ for the arithmetic progression.The $n^{th}$ term is given by $a_n = a + (n - 1)d$.For t

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$79$

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(B) Let the first term be $a$ and the common difference be $d$ for the arithmetic progression.The $n^{th}$ term is given by $a_n = a + (n - 1)d$.For the $9^{th}$ term: $a + 8d = 35$ (Equation $1$).For the $19^{th}$ term: $a + 18d = 75$ (Equation $2$).Subtracting Equation $1$ from Equation $2$: $(a + 18d) - (a + 8d) = 75 - 35$.$10d = 40$,so $d = 4$.Substituting $d = 4$ into Equation $1$: $a + 8(4) = 35 \implies a + 32 = 35 \implies a = 3$.The $20^{th}$ term is $a_{20} = a + 19d$.$a_{20} = 3 + 19(4) = 3 + 76 = 79$.

If the $9^{th}$ and $19^{th}$ terms of an arithmetic progression are $35$ and $75$ respectively,what is its $20^{th}$ term?

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