Question

tìm a
a + (a+1) + (a+2) + (a+3) + ....... (a+19)= 590

Answer 1

a+(a+1)+(a+2)+...+(a+19)=590

a+a+1+a+2+...+a+19=590

20a+(1+2+3+...+19)=590

20a+190=590

20a=400

a=20

Answer 2

a + (a+1) + (a+2) + (a+3) + ....... (a+19)= 590

⇒ (a+a+...+a)+(1+2+...+19)=590

⇒ 20a+190=590

⇒ 20a=400

⇒ a=20

Answer 3

Ta có: \(a+\left(a+1\right)+\left(a+2\right)+\left(a+3\right)+...+\left(a+19\right)=590\)

\(\Leftrightarrow20a+190=590\)

\(\Leftrightarrow20a=400\)

hay a=20

Answer 4

\(a+\left(a+1\right)+\left(a+2\right)+...+\left(a+19\right)=590\)

\(20a+\left[\left(19+1\right)\cdot19\div2\right]=590\)

\(20a+190=590\)

\(20a=400\)

a=20