The sum of the first four terms of an A.P. is | vedclass

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(A) Let the $A.P.$ be $a, a+d, a+2d, \ldots, a+(n-1)d$.The sum of the first four terms is $a + (a+d) + (a+2d) + (a+3d) = 4a + 6d = 56$.Given $a = 11$,

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$11$

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(A) Let the $A.P.$ be $a, a+d, a+2d, \ldots, a+(n-1)d$.The sum of the first four terms is $a + (a+d) + (a+2d) + (a+3d) = 4a + 6d = 56$.Given $a = 11$,we have $4(11) + 6d = 56$ $\Rightarrow 44 + 6d = 56$ $\Rightarrow 6d = 12$ $\Rightarrow d = 2$.The sum of the last four terms is $[a+(n-4)d] + [a+(n-3)d] + [a+(n-2)d] + [a+(n-1)d] = 4a + (4n - 10)d = 112$.Substituting $a = 11$ and $d = 2$:$4(11) + (4n - 10)(2) = 112$$44 + 8n - 20 = 112$$8n + 24 = 112$$8n = 88$$n = 11$.Thus,the number of terms is $11$.

The sum of the first four terms of an $A.P.$ is $56$. The sum of the last four terms is $112$. If its first term is $11$,then find the number of terms.

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