Let a1, a2, a3, ldots ldots, an be in geometric progression. If a1 +a3+a5=455 and a2+a4+a6=1365 then the ratio between each consecutive term is

Question

Let a1, a2, a3, ldots ldots, an be in geometric progression. If a1 +a3+a5=455 and a2+a4+a6=1365 then the ratio between each consecutive term is (A) 2 (B) 3 (C) 4 (D) any other number

Tardigrade Admin · Accepted Answer

a+a r2+a r4=455 a r+a r3+a r5=1365 ⇒ (a r(1+r2+r4)/a(1+r2+r4))=(1365/455) ⇒ r=3

Q. Let $a_{1}, a_{2}, a_{3}, \dots\dots, a_{n}$ be in geometric progression. If $a_{1}$ $+ a_{3} + a_{5} = 455$ and $a_{2} + a_{4} + a_{6} = 1365$ then the ratio between each consecutive term is

2

3

4

any other number

$a + a r^{2} + a r^{4} = 455$
$a r + a r^{3} + a r^{5} = 1365$
$\Rightarrow \frac{a r ( 1 + r ^{2} + r ^{4} )}{a ( 1 + r ^{2} + r ^{4} )} = \frac{1365}{455}$
$\Rightarrow r = 3$

Q. Let a1​,a2​,a3​,……,an​ be in geometric progression. If a1​ +a3​+a5​=455 and a2​+a4​+a6​=1365 then the ratio between each consecutive term is

2

3

4

any other number

a+ar2+ar4=455 ar+ar3+ar5=1365 ⇒a(1+r2+r4)ar(1+r2+r4)​=4551365​ ⇒r=3

Q. Let $a_{1}, a_{2}, a_{3}, \dots\dots, a_{n}$ be in geometric progression. If $a_{1}$ $+ a_{3} + a_{5} = 455$ and $a_{2} + a_{4} + a_{6} = 1365$ then the ratio between each consecutive term is

$a + a r^{2} + a r^{4} = 455$
$a r + a r^{3} + a r^{5} = 1365$
$\Rightarrow \frac{a r ( 1 + r ^{2} + r ^{4} )}{a ( 1 + r ^{2} + r ^{4} )} = \frac{1365}{455}$
$\Rightarrow r = 3$