Arithmetic 81811

In which arithmetic sequence is the sum of the first five terms with odd indices equal to 85 and the sum of the first five terms with even indices equal to 100?

Correct answer:

a₁ =  5
d =  3

Step-by-step explanation:

$$\begin{gathered} a_1+a_3+a_5+a_7+a_9=85 \\ {a}_2+a_4+a_6+a_8+a_{10}=100 \\ 5a_1+(2+4+6+8)d=85 \\ 5a_1+(1+3+5+7+9)d=100 \\ 5\cdot a_1+(2+4+6+8)\cdot d=85 \\ 5\cdot a_1+(1+3+5+7+9)\cdot d=100 \\ 5a_1+20d=85 \\ 5a_1+25d=100 \\ Row2-Row1\to Row2 \\ 5a_1+20d=85 \\ 5d=15 \\ d=\frac{15}{5}=3 \\ {a}_1=\frac{85-20d}{5}=\frac{85-20\cdot 3}{5}=5 \\ {a}_1=5 \\ d=3 \end{gathered}$$

$$\begin{gathered} d=3 \\ \text{ Verifying Solution: } \\ {a}_2=a_1+d=5+3=8 \\ {a}_3=a_2+d=8+3=11 \\ {a}_4=a_3+d=11+3=14 \\ {a}_5=a_4+d=14+3=17 \\ {a}_6=a_5+d=17+3=20 \\ {a}_7=a_6+d=20+3=23 \\ {a}_8=a_7+d=23+3=26 \\ {a}_9=a_8+d=26+3=29 \\ {a}_{10}=a_9+d=29+3=32 \\ {s}_1=a_1+a_3+a_5+a_7+a_9=5+11+17+23+29=85 \\ {s}_2=a_2+a_4+a_6+a_8+a_{10}=8+14+20+26+32=100 \end{gathered}$$

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