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Question Number 110016 by ZiYangLee last updated on 26/Aug/20

The solution of the equation (x+1)+(x+4)+(x+7)+...+(x+28)=155 is given by x = _____.

$$\mathrm{The}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left({x}+\mathrm{1}\right)+\left({x}+\mathrm{4}\right)+\left({x}+\mathrm{7}\right)+...+\left({x}+\mathrm{28}\right)=\mathrm{155} \\ $$$$\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:{x}\:=\:\_\_\_\_\_. \\ $$

Answered by Aziztisffola last updated on 26/Aug/20

10x+1+4+7+...+28=155 10x+((1+28)/2)×10=155 10x+5×29=155 10x+145=155 10x=10 ⇒x=1

$$\:\mathrm{10}{x}+\mathrm{1}+\mathrm{4}+\mathrm{7}+...+\mathrm{28}=\mathrm{155} \\ $$$$\:\mathrm{10}{x}+\frac{\mathrm{1}+\mathrm{28}}{\mathrm{2}}×\mathrm{10}=\mathrm{155} \\ $$$$\:\mathrm{10}{x}+\mathrm{5}×\mathrm{29}=\mathrm{155} \\ $$$$\:\mathrm{10}{x}+\mathrm{145}=\mathrm{155} \\ $$$$\:\mathrm{10}{x}=\mathrm{10}\:\Rightarrow{x}=\mathrm{1} \\ $$

Commented by aurpeyz last updated on 26/Aug/20

pls how did you get 10x in the first line?

$$\mathrm{pls}\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\mathrm{10x}\:\mathrm{in}\:\mathrm{the}\:\mathrm{first}\:\mathrm{line}? \\ $$

Commented by Aziztisffola last updated on 26/Aug/20

x+x+...+x=10x 1+ 4+...+28_(⌣_(10 terms) )

$$\mathrm{x}+\mathrm{x}+...+\mathrm{x}=\mathrm{10x} \\ $$$$\underset{\underset{\mathrm{10}\:\mathrm{terms}} {\smile}} {\mathrm{1}+\:\mathrm{4}+...+\mathrm{28}} \\ $$

Answered by som(math1967) last updated on 26/Aug/20

x+x+....10term+(1+4+..+28)=155 10x+((10)/2){2×1+(10−1)×3}=155 ⇒10x+145=155 ∴x=((10)/(10))=1ans

$$\mathrm{x}+\mathrm{x}+....\mathrm{10term}+\left(\mathrm{1}+\mathrm{4}+..+\mathrm{28}\right)=\mathrm{155} \\ $$$$\mathrm{10x}+\frac{\mathrm{10}}{\mathrm{2}}\left\{\mathrm{2}×\mathrm{1}+\left(\mathrm{10}−\mathrm{1}\right)×\mathrm{3}\right\}=\mathrm{155} \\ $$$$\Rightarrow\mathrm{10x}+\mathrm{145}=\mathrm{155} \\ $$$$\therefore\mathrm{x}=\frac{\mathrm{10}}{\mathrm{10}}=\mathrm{1ans} \\ $$

Answered by floor(10²Eta[1]) last updated on 26/Aug/20

notice that (1, 4, 7, .., 28) is an AP with ratio=3 a_n =a_1 +(n−1)r 28=1+(n−1)3⇒n=10 sum of the terms of a AP is (((a_1 +a_n )n)/2)=(((1+28).10)/2)=145 ∴ (x+1)+(x+4)+(x+7)+...+(x+28) =10x+145=155⇒10x=10⇒x=1

$$\mathrm{notice}\:\mathrm{that}\:\left(\mathrm{1},\:\mathrm{4},\:\mathrm{7},\:..,\:\mathrm{28}\right)\:\mathrm{is}\:\mathrm{an}\:\mathrm{AP}\:\mathrm{with}\: \\ $$$$\mathrm{ratio}=\mathrm{3} \\ $$$$\mathrm{a}_{\mathrm{n}} =\mathrm{a}_{\mathrm{1}} +\left(\mathrm{n}−\mathrm{1}\right)\mathrm{r} \\ $$$$\mathrm{28}=\mathrm{1}+\left(\mathrm{n}−\mathrm{1}\right)\mathrm{3}\Rightarrow\mathrm{n}=\mathrm{10} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{a}\:\mathrm{AP}\:\mathrm{is} \\ $$$$\frac{\left(\mathrm{a}_{\mathrm{1}} +\mathrm{a}_{\mathrm{n}} \right)\mathrm{n}}{\mathrm{2}}=\frac{\left(\mathrm{1}+\mathrm{28}\right).\mathrm{10}}{\mathrm{2}}=\mathrm{145} \\ $$$$\therefore\:\left(\mathrm{x}+\mathrm{1}\right)+\left(\mathrm{x}+\mathrm{4}\right)+\left(\mathrm{x}+\mathrm{7}\right)+...+\left(\mathrm{x}+\mathrm{28}\right) \\ $$$$=\mathrm{10x}+\mathrm{145}=\mathrm{155}\Rightarrow\mathrm{10x}=\mathrm{10}\Rightarrow\mathrm{x}=\mathrm{1} \\ $$

Commented by aurpeyz last updated on 26/Aug/20

how is number of terms equal to 10?

$${how}\:{is}\:{number}\:{of}\:{terms}\:{equal}\:{to}\:\mathrm{10}? \\ $$

Commented by floor(10²Eta[1]) last updated on 27/Aug/20

3 and 4 line is the formula for the general term of an AP

$$\mathrm{3}\:\mathrm{and}\:\mathrm{4}\:\mathrm{line}\:\mathrm{is}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{the}\: \\ $$$$\mathrm{general}\:\mathrm{term}\:\mathrm{of}\:\mathrm{an}\:\mathrm{AP} \\ $$

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