Let S_n denote the sum of first n terms of an arithmetic progression. If S₂₀ = 790 and S₁₀ = 145, then S₁₅ - S₅ is :

The correct option is 395.

Let the first term of the arithmetic progression (AP) be a and the common difference be d.
The sum of the first n terms of an AP, denoted by S_n, is given by the formula:

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

We are given two equations:
1) S₂₀ = 790
2) S₁₀ = 145

Let us write the equations using the formula:

For S₂₀ = 790:
$\frac{20}{2}[2a+(20-1\left)d\right]=790$
$10[2a+19d]=790$
$2a+19d=79$ ---- (Equation 1)

For S₁₀ = 145:
$\frac{10}{2}[2a+(10-1\left)d\right]=145$
$5[2a+9d]=145$
$2a+9d=29$ ---- (Equation 2)

Now, subtract Equation 2 from Equation 1 to find the common difference d:
$(2a+19d)-(2a+9d)=79-29$
$10d=50$
$d=5$

Substitute the value of d = 5 back into Equation 2 to find a:
$2a+9\left(5\right)=29$
$2a+45=29$
$2a=29-45$
$2a=-16$
$a=-8$

Now, we need to calculate S₁₅ - S₅:
Using the formula for S_n:

$S_{15}=\frac{15}{2}\left[2\right(-8)+14(5\left)\right]$
$S_{15}=\frac{15}{2}[-16+70]$
$S_{15}=\frac{15}{2}\left[54\right]=15\times 27=405$

$S_5=\frac{5}{2}\left[2\right(-8)+4(5\left)\right]$
$S_5=\frac{5}{2}[-16+20]$
$S_5=\frac{5}{2}\left[4\right]=5\times 2=10$

Therefore, we calculate the difference:
$S_{15}-S_5=405-10=395$