Let a, b, c and d are in a geometric progression such that a

Question

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

MARKS Content Team · Accepted Answer

Correct answer: (D) 1456. Let r be the common ratio a+d=112⇒a+ar^3=112 b+c=48⇒ar+ar^2=48 Dividing the first equation by the second equation we get (a1+r^3)/(a1+rr)=112/48=(1+r1-r+r^2)/(rr+1)=7/3 ⇒3r^2-3r+3=7r ⇒3r^2-10r+3=0 r=3 or 1/3 Given they are in ascending order ⇒r=3 r=3⇒a=112/(1+r^3)=112/28=4 ⇒a=4,b=12,c=36,d=108 ⇒ Sum of first 6 terms =(4(3^6-1))/((3-1))=1456

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

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