The solution of the equation (x + 1) + (x + 4 | vedclass

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(A) The given equation is $(x + 1) + (x + 4) + (x + 7) + \dots + (x + 28) = 155$.This is an arithmetic progression $(A.P.)$ where the first term $a =

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(A) The given equation is $(x + 1) + (x + 4) + (x + 7) + \dots + (x + 28) = 155$.This is an arithmetic progression $(A.P.)$ where the first term $a = (x + 1)$ and the common difference $d = 3$.Let $n$ be the number of terms. The last term $l = (x + 28)$.The formula for the $n^{th}$ term is $l = a + (n - 1)d$.Substituting the values: $(x + 28) = (x + 1) + (n - 1)3$.$27 = (n - 1)3 \Rightarrow n - 1 = 9 \Rightarrow n = 10$.The sum of an $A.P.$ is given by $S_n = \frac{n}{2}(a + l)$.Substituting the values: $155 = \frac{10}{2}[(x + 1) + (x + 28)]$.$155 = 5(2x + 29)$.$31 = 2x + 29$.$2x = 2 \Rightarrow x = 1$.

The solution of the equation $(x + 1) + (x + 4) + (x + 7) + \dots + (x + 28) = 155$ is

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