d = 3 , a₁₂ = 40 and S[tex]x_{100}[/tex] = 7775
In an arithmetic sequence the nth term and sum to n terms are
• a[tex]n[/tex] = a₁ + (n-1)d
• S[tex]_n[/tex] = [tex]\frac{n}{2}[/tex][2a + (n-1)d]
where d is the common difference
a₆ = a₁ + 5d = 22 ⇒ 7 + 5d = 22 ⇒ 5d = 15 ⇔ d = 3
a₁₂ = 7 + 11d = 7 +( 11× 3) = 7 + 33 = 40
S₁₀₀ = [tex]\frac{50}{2}[/tex][(2×7) +(99×3)
= 25(14 + 297) = 25(311)= 7775
