Question

Calculati:
a) 2+5+8+...+242=
b) 3+7+11+...+43=

Answer 1

$\displaystyle a).2+5+8+...+242 \\ \\ 242=2+(n-1) \times 3 \\ \\ 242=2+3n-3 \\ \\ 3n=242-2+3 \\ \\ 3n=243 \\ \\ n= \frac{243}{3} \\ \\ n=81$

$\displaystyle S_{81}= \frac{4+80 \times 3}{2} \times 81 \\ \\ S_{81}= \frac{4+240}{2} \times 81 \\ \\ S_{81}= \frac{244}{2} \times 81 \\ \\ S_{81}= 122 \times 81 \\ \\ S_{81}=9882$

$\displaystyle b).3+7+11+...+43 \\ \\ 43=3+(n-1) \times 4 \\ \\ 43=3+4n-4 \\ \\ 4n=43-3+4 \\ \\ 4n=44 \\ \\ n= \frac{44}{4} \\ \\ n=11$

$\displaystyle S_1_1= \frac{6+10 \times 4}{2} \times 11 \\ \\ S_1_1= \frac{6+40}{2} \times 11 \\ \\ S_1_1= \frac{46}{2} \times 11 \\ \\ S_1_1=23 \times 11 \\ \\ S_1_1=253$