' n ' arithmetic means A1, A2 ldots, An are inserted between 7 and 49 such that sum of these means is 364 . If the G.M. of A3 and A6 is k, then ((k/4)) is equal to

Question

' n ' arithmetic means A1, A2 ldots, An are inserted between 7 and 49 such that sum of these means is 364 . If the G.M. of A3 and A6 is k, then ((k/4)) is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

We have A1+A2+ ldots+An=n((7+49/2))=364 ⇒ n=13 Hence 49=7+14 d ⇒ d=3 ∴ A1=7+3=10, A2=13, A3=16, ldots, A6=25 ⇒ k=√16.25=20 .

Q. ' $n$ ' arithmetic means $A_{1}, A_{2} \dots, A_{n}$ are inserted between $7$ and $49$ such that sum of these means is $364$ . If the $G . M$ . of $A_{3}$ and $A_{6}$ is $k$ , then $(\frac{k}{4})$ is equal to

We have $A_{1} + A_{2} + \dots + A_{n} = n (\frac{7 + 49}{2}) = 364$
$\Rightarrow n = 13$
Hence $49 = 7 + 14 d \Rightarrow d = 3$
$∴ A_{1} = 7 + 3 = 10, A_{2} = 13, A_{3} = 16, \dots, A_{6} = 25$
$\Rightarrow k = 16.25 = 20.$

Q. ' n ' arithmetic means A1​,A2​…,An​ are inserted between 7 and 49 such that sum of these means is 364 . If the G.M. of A3​ and A6​ is k, then (4k​) is equal to

We have A1​+A2​+…+An​=n(27+49​)=364 ⇒n=13 Hence 49=7+14d⇒d=3 ∴A1​=7+3=10,A2​=13,A3​=16,…,A6​=25 ⇒k=16.25​=20.

Q. ' $n$ ' arithmetic means $A_{1}, A_{2} \dots, A_{n}$ are inserted between $7$ and $49$ such that sum of these means is $364$ . If the $G . M$ . of $A_{3}$ and $A_{6}$ is $k$ , then $(\frac{k}{4})$ is equal to

We have $A_{1} + A_{2} + \dots + A_{n} = n (\frac{7 + 49}{2}) = 364$
$\Rightarrow n = 13$
Hence $49 = 7 + 14 d \Rightarrow d = 3$
$∴ A_{1} = 7 + 3 = 10, A_{2} = 13, A_{3} = 16, \dots, A_{6} = 25$
$\Rightarrow k = 16.25 = 20.$