Let a, b, c and d are in a geometric progression such that `a lt b lt c lt d, a + d=112` amd `b+c=48`. If the geometric progression is continued with a as the first term, then the sum of the first six terms is

a,b,c,d are in GP are in ascending order such that a+d=112 and b+c=48 If the GP is continued with as the first term, then the sum of the first six term is:

The sum of an infinite geometric progression is 6, If the sum of the first two terms is 9//2 , then what is the first term?

If a,b,c are the first three non-zero terms of a geometric progression such that a-2,2b and 12c forms another geometric progression with common ratio 5, then the sum of the series a+b+c+...+oo, is

The third term of a geometric progression is 4. Then find the product of the first five terms.

The third term of a geometric progression is 4. Then find the product of the first five terms.

The third term of a geometric progression is 4. Then find the product of the first five terms.

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 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