Quick Sum

We were given that $x = 100 + 99+98+\cdots+50$

$\begin{array}{c} ~ & 100 &+&99 &+& 98 &+& 97 &+& \cdots &+& 50\\ + & 50 &+&51 &+& 52 &+& 53 &+& \cdots &+& 100\\ \hline ~ & 150 &+& 150 &+& 150 &+& 150 &+& \cdots &+& 150 \end{array}$

So $2x = 51(150)~~~\Longrightarrow~~~x = \dfrac{51(150)}{2} = \boxed{3825}$

sum of arithmetic progressions has this formula

Sn=(a+l)n/2,

where n is the number of terms, a is the first term and l is the last term

According to the question, there are 51 terms, with the first term being 50 and the last term being 100

so Sn=51 x (50+100)/2 =51 x 75 =3825

$x = (100 + 99 + \ldots+ 1) - (49+48+\ldots+1) = 5050 - 1225 = 3825$

Because $1 + 2 + 3 + \ldots + n = \frac12 n(n+1)$ .

I feel the easiest way is the find the average and then multiply it by the number of digits. Here the average is easy to find, i.e. 75. Then we know that from 50 to 100 there are 51 digits and the answer to their total is 75 * 51 = 3825

By using this formula:

$S_{n}=\frac{n}{2} (U_{n}+a)$

to find n we can use:

if $n=1$ then the number is $50$

so, $U_{1}=50$ , $U_{n}=50-1=49'$

$U_{n}=100, 49'=100, n=100-49=51$

Subtitude:

$S_{51}=\frac{51}{2} (100+50)$

$S_{51}=\frac{51}{2} (150)=51 \times 75= 3825$

We can do this question by sum of arithmetic progression formula.. Sn=n/2(a+an)....where a=first term and an=last term and n=no. of term

If we take these series like this.. 50,51,52.........98,99,100 Then here, a=50,common difference(d)=1 an=100 And an=a+(n-1)d,therefore.. 100=50+(n-1)1 n=51 So, Now, Sn=51/2(50+100) Sn=3825(answer)

1+2+3+.......+100=(100×101)/2=5050 1+2+3+......+49=(49×50)/2=1225

. '. 50+51+52+.....100=5050-1225=3825

Equation=n×(n+1)/2

sum of n numbers = n(n+1)/2. so for first 49 numbers 49(49+1)/2 = 25 49. for first 100 numbers 100(100+1)/2 = 50 101 Then subtract 50(101)-25(49) = 3825

Using the formula $\displaystyle \sum_{k=a}^{a+n} k=\frac{(n+1)(2a+n)}{2}$ as shown in this problem

So $\frac{(2 \times 50+(100-50))((100-50)+1)}{2}=\frac{150 \times 51}{2}=\boxed{\large{3825}}$

$d=-1$

$a_n=a_1+(n-1)(d)$

$50=100+(n-1)(-1)$

$n=51$

$S=$ $\frac{n}{2}$ $[(2a_1+(n-1)(d)]$ $=$ $\frac{51}{2}$ $[2(100)+(50)(-1)]=$ $3825$

Subtract the sum of 1st 49 terms from

Sum of 1st 100 terms

5050-1225=3825

Since $\large \sum_{i=1}^{n}i=\frac12 n(n+1)$

Therefore the given sum will be $\frac{100(100+1)}{2} - \frac{49(49+1)}{2}$

$=5050-1225$

$=3825$

There are a number of ways to approach this problem.