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 where the first term $a = x + 1$,th

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

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

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