Omitted Progression

$\text{We have the formula of sum to n terms of an Arithmetic Progression with first and last terms as 'a' and 'l' respectively:-}$ $\dfrac{n}2(a+l)$ .

$\text{So,by the formula,the sum of all the terms of the given A.P. is 2500.}$

$\text{Also, the sum of the first and last terms is 1000(i.e.,1+999).}$ $\text{Therefore the sum of all the missing terms is}$ $2500-1000=1500$ .

Relevant wiki: Arithmetic Progressions

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Let the common difference for this A.P. be $d$ .

Now, sum of first and last terms $= T_1 + T_5 = 1 + 999 = 1000$ .

Sum of second and second last terms $= T_2 + T_4 = (1+d) + (999-d) = 1000$ .

Sum of third and third last terms $= T_3 + T_3 =(1+2d) + (999-2d) = 1000 \implies 2 \times T_3 = 1000 \implies T_3 = 500$ .

Therefore, required sum: $(T_2 + T_4) + T_3 \implies 1000 + 500 = \boxed{1500}$

Relevant wiki: Arithmetic Progressions

$\large 1, \ x, \ y, \ z, 999$

Let the first term and common difference of AP be $a$ and $d$ .

We know that :

$\implies a_{n^{th}}=a+(n-1)d$

And,

$\implies a_1=a+0×d=1$

$\implies a_5=a+4d=999$

$d=\dfrac{998}{4}=249.5$

$\therefore x+y+z=(a+d)+(a+2d)+(a+3d)=3a+6d=3×1+6×249.5=\boxed{1500}$

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$\large \mathcal {\color{#3D99F6}{Alternate \ solution}}$

$\large 1, \ x, \ y, \ z, 999$

Using Arithematic mean:

$\implies \dfrac{1+999}{2}=y$

$\color{#D61F06}{y}=500 \ ...(1)$

$\implies \dfrac{1+\color{#D61F06}{y}}{2}=x \ ...(2)$

$\implies \dfrac{\color{#D61F06}{y}+999}{2}=z \ ...(3)$

$(1)+(2)+(3)$

$x+y+z=500+\dfrac{1000+2\color{#D61F06}{y}}{2}=500+1000=\boxed{1500}$

Because the progression is arithmetic there is a constant difference, a, between any two adjacent terms. Thus it can be written as 1, 1+a, 1+2a, 1+3a, 1+4a,.... So 999=1+4a, one can then solve for a=249.5. And the sum of the middle three terms is 3+6(249.5)=1500