1.Suppose 28−p28 - p28−p, ppp, 70−q70 - q70−q, qqq are the coefficients of four consecutive terms in the expansion of (1+x)n(1+x)^{n}(1+x)n. Then the value of 2q−3p2q - 3p2q−3p equals:a.777b.101010c.444d.666Login to continueOnly logged in users canattempt or see the solution.