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 $a = 11$.Sum of the first four terms: $a + (a+d) + (a+2d) + (a+3d) = 56$.

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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 $a = 11$.Sum of the first four terms: $a + (a+d) + (a+2d) + (a+3d) = 56$.$4a + 6d = 56 \implies 4(11) + 6d = 56 \implies 44 + 6d = 56 \implies 6d = 12 \implies d = 2$.Sum of the last four terms: $(a+(n-1)d) + (a+(n-2)d) + (a+(n-3)d) + (a+(n-4)d) = 112$.$4a + (4n - 10)d = 112$.Substitute $a = 11$ and $d = 2$: $4(11) + (4n - 10)(2) = 112$.$44 + 8n - 20 = 112 \implies 8n + 24 = 112 \implies 8n = 88 \implies n = 11$.

The sum of the first four terms of an arithmetic progression is $56$ and the sum of its last four terms is $112$. If the first term is $11$,what is the number of terms in the progression?

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