Let `a_(1), a_(2), a_(3).... and b_(1), b_(2), b_(3)...` be arithmetic progression such that `a_(1) = 25, b_(1) = 75 and a_(100) + b_(100) = 100`. Then

Let a_(1),a_(2),a_(3)... and b_(1),b_(2),b_(3)... be arithmetic progressions such that a_(1)=25,b_(1)=75 and a_(100)+b_(100)=100 then the sum of first hundred terms of the progression a_(1)+b_(1)a_(2)+b_(2) is

Let a_(1), a_(2), cdots and b_(1), b_(2) cdots be arithmetic progression such that a_(1) = 25, b_(1) = 75 and a_(100) + b_(100) = 100 , then the sum of first hundred terms of the progression a_(1) + b_(1), a_(2) + b_(2), cdots is

Let a_(1), a_(2),…. and b_(1),b_(2),…. be arithemetic progression such that a_(1)=25 , b_(1)=75 and a_(100)+b_(100)=100 , then the sum of first hundred term of the progression a_(1)+b_(1) , a_(2)+b_(2) ,…. is equal to

Let a_(1), a_(2),…. and b_(1),b_(2),…. be arithemetic progression such that a_(1)=25 , b_(1)=75 and a_(100)+b_(100)=100 , then the sum of first hundred term of the progression a_(1)+b_(1) , a_(2)+b_(2) ,…. is equal to

Let a_(1), a_(2),…. and b_(1),b_(2),…. be arithemetic progression such that a_(1)=25 , b_(1)=75 and a_(100)+b_(100)=100 , then the sum of first hundred term of the progression a_(1)+b_(1) , a_(2)+b_(2) ,…. is equal to

Let a_(1), a_(2),…. and b_(1),b_(2),…. be arithemetic progression such that a_(1)=25 , b_(1)=75 and a_(100)+b_(100)=100 , then the sum of first hundred term of the progression a_(1)+b_(1) , a_(2)+b_(2) ,…. is equal to

If a_(1),a_(2),a_(3)...b_(1),b_(1),b_(2),b_(3)...... are in AP. Such that a_(1)+b_(1)=a_(100)+b_(100)=16 and sum_(x=1)^(n)r(a_(r)+b_(r))=576 the find n

If a_1,a_2,a_3,....a_n and b_1,b_2,b_3,....b_n are two arithematic progression with common difference of 2nd is two more than that of first and b_(100)=a_(70),a_(100)=-399,a_(40)=-159 then the value of b_1 is

If the arithmetic mean of a_(1),a_(2),a_(3),"........"a_(n) is a and b_(1),b_(2),b_(3),"........"b_(n) have the arithmetic mean b and a_(i)+b_(i)=1 for i=1,2,3,"……."n, prove that sum_(i=1)^(n)(a_(i)-a)^(2)+sum_(i=1)^(n)a_(i)b_(i)=nab .

If the arithmetic mean of a_(1),a_(2),a_(3),"........"a_(n) is a and b_(1),b_(2),b_(3),"........"b_(n) have the arithmetic mean b and a_(i)+b_(i)=1 for i=1,2,3,"……."n, prove that sum_(i=1)^(n)(a_(i)-a)^(2)+sum_(i=1)^(n)a_(i)b_(i)=nab .