The sum of the first four terms of an arithme | vedclass

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(B) Let the arithmetic progression be $a, a+d, a+2d, \dots, a+(n-1)d$.Given the first term $a = 11$.The sum of the first four terms is $a + (a+d) + (a

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$11$

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(B) Let the arithmetic progression be $a, a+d, a+2d, \dots, a+(n-1)d$.Given the first term $a = 11$.The sum of the first four terms is $a + (a+d) + (a+2d) + (a+3d) = 4a + 6d = 56$.Substituting $a = 11$: $4(11) + 6d = 56$ $\Rightarrow 44 + 6d = 56$ $\Rightarrow 6d = 12$ $\Rightarrow d = 2$.The sum of the last four terms is $t_{n-3} + t_{n-2} + t_{n-1} + t_n = 112$.In an arithmetic progression,the sum of terms equidistant from the beginning and end is constant: $t_1 + t_n = t_2 + t_{n-1} = t_3 + t_{n-2} = t_4 + t_{n-3} = k$.Thus,$(t_1 + t_n) + (t_2 + t_{n-1}) + (t_3 + t_{n-2}) + (t_4 + t_{n-3}) = 56 + 112 = 168$.Since each pair equals $k$,we have $4k = 168 \Rightarrow k = 42$.Therefore,$t_1 + t_n = 42$.Substituting $t_1 = 11$: $11 + t_n = 42 \Rightarrow t_n = 31$.Using the formula $t_n = a + (n-1)d$: $31 = 11 + (n-1)2$.$20 = (n-1)2$ $\Rightarrow n-1 = 10$ $\Rightarrow n = 11$.

The sum of the first four terms of an arithmetic progression is $56$. The sum of the last four terms is $112$. If its first term is $11$,then the number of terms is:

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