Correction:Given Sₙ = 2n² + 5n
S₁ = 2×1 + 5×1 = 7
S₂ = 2×(2²) + 5×2 = 8 + 10 = 18
Therfore First term a₁ = 7 & Second term a₂ = 18 - 7 = 11
and therefore common difference 'd' = 11 - 7 = 4
Now nth term
aₙ = a₁ + (n-1)d
= 7 + ( n - 1) 4
aₙ = 7 + 4n - 4
aₙ = 4n + 3
Hence Proved!