Prove that `|{:(a_(1)alpha_(1)+b_(1)beta_(1),a_(1)alpha_(2)+b_(1)beta_(2),a_(1)alpha_(3)+b_(1)beta_(3)),(a_(2)alpha_(1)+b_(2)beta_(1),a_(2)alpha_(2)+b_(2)beta_(2),a_(2)alpha_(3)+b_(2)beta_(3)),(a_(3)alpha_(1)+b_(3)beta_(1),a_(3)alpha_(2)+b_(3)beta_(2),a_(3)alpha_(3)+b_(3)beta_(3)):}|`=0.

If a_(1),a_(2),a_(3),a_(4),a_(5) are in HP, then a_(1)a_(2)+a_(2)a_(3)+a_(3)a_(4)+a_(4)a_(5) is equal to

the value of the determinant |{:((a_(1)-b_(1))^(2),,(a_(1)-b_(2))^(2),,(a_(1)-b_(3))^(2),,(a_(1)-b_(4))^(2)),((a_(2)-b_(1))^(2),,(a_(2)-b_(2))^(2) ,,(a_(2)-b_(3))^(2),,(a_(3)-b_(4))^(2)),((a_(3)-b_(1))^(2),,(a_(3)-b_(2))^(2),,(a_(3)-b_(3))^(2),,(a_(3)-b_(4))^(2)),((a_(4)-b_(1))^(2),,(a_(4)-b_(2))^(2),,(a_(4)-b_(3))^(2),,(a_(4)-b_(4))^(2)):}| is

Find the coefficient of x in the determinant |{:((1+x)^(a_(1)b_(1)),(1+x)^(a_(1)b_(2)),(1+x)^(a_(1)b_(3))),((1+x)^(a_(2)b_(1)),(1+x)^(a_(2)b_(2)),(1+x)^(a_(2)b_(3))),((1+x)^(a_(3)b_(1)),(1+x)^(a_(3)b_(2)),(1+x)^(a_(3)b_(3))):}|

If |a_(1)|gt|a_(2)|+|a_(3)|,|b_(2)|gt|b_(1)|+|b_(3)| and |c_(2)|gt|c_(1)|+|c_(2)| then show that |{:(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3)):}|ne0.

If a_(1),a_(2),".....", are in HP and f_(k)=sum_(r=1)^(n)a_(r)-a_(k) , then 2^(alpha_(1)),2^(alpha_(2)),2^(alpha_(3))2^(alpha_(4)),"....." are in {" where " alpha_(1)=(a_(1))/(f_(1)),alpha_(2)=(a_(2))/(f_(2)),alpha_(3)=(a_(3))/(f_(3)),"....."} .

If alpha_(1), alpha_(2), alpha_(3), beta_(1), beta_(2), beta_(3) are the values of n for which sum_(r=0)^(n-1)x^(2r) is divisible by sum_(r=0)^(n-1)x^(r ) , then the triangle having vertices (alpha_(1), beta_(1)),(alpha_(2),beta_(2)) and (alpha_(3), beta_(3)) cannot be

Let -pi/6 beta_1 and alpha_2 >beta_2 , then alpha_1 + beta_2 equals:

if a_(1)b_(1)c_(1), a_(2)b_(2)c_(2)" and " a_(3)b_(3)c_(3) are three-digit even natural numbers and Delta = |{:(c_(1),,a_(1),,b_(1)),(c_(2),,a_(2),,b_(2)),(c_(3),,a_(3),,b_(3)):}|" then " Delta is

If a_(1),a_(2),a_(3),"........",a_(n) are in AP with a_(1)=0 , prove that (a_(3))/(a_(2))+(a_(4))/(a_(3))+"......"+(a_(n))/(a_(n-1))-a_(2)((1)/(a_(2))+(1)/(a_(3))"+........"+(1)/(a_(n-2))) = (a_(n-1))/(a_(2))+(a_(2))/(a_(n-1)) .

Prove that |[[alpha,beta,gamma],[alpha^2,beta^2,gamma^2],[alpha^3,beta^3,gamma^3]]|= alpha beta gamma (alpha-beta)(beta-gamma)(gamma-alpha)