For the A.P. 4, 8, 12, 16, ldots,T_{40} - T_{ | vedclass

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(B) For an Arithmetic Progression $(A.P.)$,the $n^{th}$ term is given by $T_n = a + (n - 1)d$.Here,the first term $a = 4$ and the common difference $d

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$40$

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(B) For an Arithmetic Progression $(A.P.)$,the $n^{th}$ term is given by $T_n = a + (n - 1)d$.Here,the first term $a = 4$ and the common difference $d = 8 - 4 = 4$.We need to find $T_{40} - T_{30}$.$T_{40} = a + (40 - 1)d = a + 39d$.$T_{30} = a + (30 - 1)d = a + 29d$.Subtracting the two terms:$T_{40} - T_{30} = (a + 39d) - (a + 29d) = 10d$.Substituting the value of $d = 4$:$T_{40} - T_{30} = 10 	imes 4 = 40$.

For the $A.P.$ $4, 8, 12, 16, \ldots$,$T_{40} - T_{30} = \ldots$

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