`b_(1) ,b_(2),b_(3)` form a geometric progression such that their product is `64` and their arithmetic mean is `(14)/(3)` .If `b_(1)>b_(2)>b_(3)` then `b_(3)`:

The first term of a geometric progression is equal to b-2, the third term is b+6, and the arithmetic mean of the first and third term to the second term is in the ratio 5:3. Find the positive integral value of b .

Let a_(1),a_(2),a_(3)..... be an arithmetic progression and b_(1),b_(2),b_(3)...... be a geometric progression.The sequence c_(1),c_(2),c_(3),... is such that c_(n)=a_(n)+b_(n)AA n in N. suppose c_(1)=1.c_(2)=4.c_(3)=15 and c_(4)=2. The common ratio of geometric progression is equal to

Let (a_(1),b_(1)) and (a_(2),b_(2)) are the pair of real numbers such that 10,a,b,ab constitute an arithmetic progression. Then, the value of ((2a_(1)a_(2)+b_(1)b_(2))/(10)) is

If b_(1),b_(2),b_(3) are three consecutive terms of an arithmetic progression with common difference dgt0, then what is the value of d for which b(2)/(3)=b_(2)b_(3)+b_(1)d+2?

Let three positive numbers a, b c are in geometric progression, such that a, b+8 , c are in arithmetic progression and a, b+8, c+64 are in geometric progression. If the arithmetic mean of a, b, c is k, then (3)/(13)k is equal to

let a < b , if the numbers a , b and 12 form a geometric progression and the numbers a , b and 9 form an arithmetic progression , then (a+b) is equal to:

Let a_(1), a_(2), a_(3)... be a sequance of positive integers in arithmetic progression with common difference 2. Also let b_(1),b_(2), b_(3)...... be a sequences of posotive intergers in geometric progression with commo ratio 2. If a_(1)=b_(1)=c_(2) . then the number of all possible values of c, for which the equality. 2(a_(1)+a_(2).+.....+a_(n))=b_(1)+b_(2)+....+b_(n) holes for same positive integer n, is

If a,b,c are in geometric progression and a,2b,3c are in arithmetic progression, then what is the common ratio r such that 0ltrlt1 ?

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