History of Hindu Mathematics: A Source Book v2.c17.s7.13

v2.c17.s7.13

*Rationale.* By the process of elimination we get from equations (I) $(a₂ - a₁) + (a₄ - a₃) + + (a_n-1 - a_n-2) + 2x₁ = a_n.$ Assume x₁ = p; so that $(a₂ - a₁) + (a₄ - a₃) + + (a_n-1 - a_n-2) + 2p = b, say.$ Subtracting 2(x₁ - p) = a_n - b. Therefore x₁ = p + ½(a_n - b).