Arithmetic progressions 2

Let the first term be $a$ , and the common difference be $d$

$T_1=a\\ T_2=a+d\\ T_3=a+2d\\ \ldots\\ T_8=a+7d\\ T_9=a+8d\\ T_{10}=a+9d$

Given that $T_1+T_2+T_3=321$ and $T_8+T_9+T_{10}=405$ , we get:

$a+a+d+a+2d=321 \implies 3a+3d=321$ ......(1)

$a+7d+a+8d+a+9d=405 \implies 3a+24d=405$ ......(2)

Eq. (2) - Eq. (1):

$(3a+24d)-(3a+3d) = 405-321\\ 21d=84 \implies d=4$

Use the value found to calculate $a$ :

$3a+3(4)=321\\ 3a=309 \implies a=103$

Then, calculate $T_{10}$ :

$T_{10}=103+9(4)=139$

Finally, calculate the sum of the $10$ numbers:

$S_{10}=\dfrac{10}{2}(103+139)=5(242)=\boxed{1210}$