An algebra problem by A A

Sum of consecutive numbers: $\frac{n(n+1)}{2}$

$1+2+3+4+5+6+7+8+9+10=\frac{10\times11}{2}=\frac{110}{2}=55$

Thus, the answer is $\boxed{55}$

(n/2)(n+1) this is the basic rule to solve this problem rather than bullying yourself by adding those numbers. (10/2)(10+1) = 5*11 = 55

10+(9+1)+(8+2)+(7+3)+(6+4)+5 =10+10+10+10+10+5 =55

sum of consecutive integer is n*(n+1)/2

$1+10=11$

$2+9=11$

.................

$5+6=11$

so, $11 \times 5=55$

The answer is $\dfrac{10 × 11}{2} = \color{#69047E}{\boxed{55}}$ .

If you see a pattern, you could see this...

1,2,3,4,5,6,7,8,9,10

1+10=11 2+9=11 3+8=11 4+7=11 5+6=11

These 11's occur 5 times Thus, 5×11=55

1+2+3+4=10, 10+5=15, 15+6=21, 21+7=28, 28+8=36, 36+9=45, 45+10=55

n(n+1)/2 =) 10 *11/2 =110/2=55 ie the answer

1+2+3+4+5+6+7+8+9+10=55

1+2+3+...n= $n\times \frac { n }{ 2 } +\frac { n }{ 2 }$ for n=10; sum= $10\times \frac { 10 }{ 2 } +\frac { 10 }{ 2 } =10\times 5+5$ =55

$sum_{r=1}^n$ r=(r(r+1))/2

10*11/2=55

10 * 11 =110 then 110 / 2

You can either use formula of sum of an A. P. or just add them up. By formula-

n\2 *[2a+(n-1)d]

where n is number of terms in A.P ,d is common difference and 'a' is first term. here, n=2 d=1 and a=1