The tenth term of an arithmetic progression is equal to twice the fourth term. The twentieth term of the progression is 44.

(i) Find the first term and the common difference.

(ii) Find the sum of the first 50 terms.

The formula for Arithmetic Progression is $u_{r}=a+\left( r-1\right) d$

where $u_{r}$ = the rth term, a = the first term and d = the common difference

Now the tenth term is equal to twice the fourth term in this arithmetic progression

$\therefore u_{10}=2u_{4}$

$u_{10}=a+\left( 10-1\right) d$ or $u_{10}=a+9d$

$u_{4}=a+\left( 4-1\right) d$ or $u_{4}=a+3d$

$\therefore a+9d=2(a+3d)$

or, $a+9d=2a+6d$

$\therefore 2a-2=9d-6d$, or $(a=3d)$

Now the 20th term $u_{20}=a+\left( 20-1\right) d=44$ or $a+19d=44$

Substituting $a=3d$, we get $3d+19d=22$ or $22d=44$ or $\underline{d=2}$

$\therefore a=3\times2$ or $\underline{a=6}$

$\therefore$ the first term a=6, and the common difference d=2

The formula for Sum of Arithmetic Progression is $S_{n}=\dfrac {1}{2}\times n\left[ 2a+\left( n-1\right) d\right]$
$\therefore S_{50}=\dfrac {1}{2}\times 50\left[ 2\times 6+\left( 50-1\right) \times 2\right]$
$S_{50}=\dfrac {1}{2}\times 50\times (12+49 \times 2) = 2750$
$\therefore$ the sum of the first 50th term is = 2750