Let a1, a2, a3, a4 are positive numbers which are consecutive terms in a geometric progression. If a1+a2=10 and a3+a4=40, then the value of (a1+a4) equals

Question

Let a1, a2, a3, a4 are positive numbers which are consecutive terms in a geometric progression. If a1+a2=10 and a3+a4=40, then the value of (a1+a4) equals (A) 30 (B) 28 (C) 26 (D) 24

Tardigrade Admin · Accepted Answer

a1(1+r)=10 ...(1) a1 r2(1+r)=40....(2) ∴ a1=(10/3), r=2

Q. Let $a_{1}, a_{2}, a_{3}, a_{4}$ are positive numbers which are consecutive terms in a geometric progression. If $a_{1} + a_{2} = 10$ and $a_{3} + a_{4} = 40$ , then the value of $(a_{1} + a_{4})$ equals

30

28

26

24

$a_{1} (1 + r) = 10$ ...(1)
$a_{1} r^{2} (1 + r) = 40$ ....(2)
$∴ a_{1} = \frac{10}{3}, r = 2$

Q. Let a1​,a2​,a3​,a4​ are positive numbers which are consecutive terms in a geometric progression. If a1​+a2​=10 and a3​+a4​=40, then the value of (a1​+a4​) equals

30

28

26

24

a1​(1+r)=10 ...(1) a1​r2(1+r)=40....(2) ∴a1​=310​,r=2

Q. Let $a_{1}, a_{2}, a_{3}, a_{4}$ are positive numbers which are consecutive terms in a geometric progression. If $a_{1} + a_{2} = 10$ and $a_{3} + a_{4} = 40$ , then the value of $(a_{1} + a_{4})$ equals

$a_{1} (1 + r) = 10$ ...(1)
$a_{1} r^{2} (1 + r) = 40$ ....(2)
$∴ a_{1} = \frac{10}{3}, r = 2$