If the 9^{th} term of an A.P. is 35 and the 1 | vedclass

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(B) The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n-1)d$.Given $T_9 = a + 8d = 35$ (Equation $1$).Given $T_{19} = a + 18d = 75$ (Equation $2$

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$79$

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(B) The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n-1)d$.Given $T_9 = a + 8d = 35$ (Equation $1$).Given $T_{19} = a + 18d = 75$ (Equation $2$).Subtracting Equation $1$ from Equation $2$:$(a + 18d) - (a + 8d) = 75 - 35$$10d = 40$$d = 4$.Substituting $d = 4$ into Equation $1$:$a + 8(4) = 35$$a + 32 = 35$$a = 3$.The $20^{th}$ term is $T_{20} = a + 19d$.$T_{20} = 3 + 19(4) = 3 + 76 = 79$.

If the $9^{th}$ term of an $A.P.$ is $35$ and the $19^{th}$ term is $75$,then its $20^{th}$ term will be:

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