Let an be a G-P such that a4-a6-1-4 and a2-a5-216- Then a1-

Suppose $a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$ are integers such that $\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4!} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}$ where 0 $\leq$ a< j for j= 2,4,5,6,7.

The sum $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$ is

Q. Suppose

a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}

are intagers such that

\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4!} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}

where

0 \leq a_{j} < j

for

j = 2, 3, 4, 5, 6, 7

. The sum

a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}

is-

Q. Let

a_{1}, a_{2}, a_{3} a_{4}, a_{5}

be a

G . P

of positive real numbers such that the

A . M

a_{2}

and

a_{4}

117

and the

G . M

a_{2}

and

a_{4}

108

. Then the

A . M

a_{1}

and

a_{5}

is:

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Let an be a G.P such that a4a6=14 and a2+a5=216. Then a1=

The correct option is A 12 or 1087Let a,ar,ar2,... be in G.P such that a4a6=14.⇒a1r3a1r5=14⇒a1r3a1r5=14⇒r=±2Case (1): When r=2a2+a5=216⇒a1r+a1r4=216⇒2a1+16a1=216⇒18a1=216⇒a1=12Case (2): When r=−2a2+a5=216⇒a1r+a1r4=216⇒−2a1+16a1=216⇒14a1=216⇒a1=21614⇒a1=1087

Let $a_{n}$ be a G.P such that $\frac{a_{4}}{a_{6}} = \frac{1}{4}$ and $a_{2} + a_{5} = 216$ . Then $a_{1} =$