Question

Arithmetic mean between two numbers is 5 and geometric mean between them is 4. Find the harmonic mean between the numbers. And in a harmonic progression third term and fifth terms are respectively 1 and \[\dfrac{-1}{5}\]. Find the tenth term.

Answer 1

Hint: As arithmetic mean is given 5 so from this information we can find the sum of two numbers which is 10. Then we will assume one number to be x and another to be 10-x.Geometric mean between two numbers are given ,So substituting the assumed values and equating it to geometric mean formula we get an quadratic equation and calculate the values of x and determine harmonic mean by using formula.From the definition of harmonic progression equate the terms 1 and \[\dfrac{-1}{5}\] with general form and simplify it and next calculate the value of tenth term

Complete step-by-step answer:
Arithmetic mean between two numbers \[=\dfrac{x+y}{2}.......(1)\]
Geometric mean between two numbers \[=\sqrt{x\times y}..........(2)\]
Here it is given arithmetic mean is 5, so from equation (1) we get,
 \[\,\Rightarrow \dfrac{x+y}{2}=5........(3)\]
Solving equation (3) we get,
 \[\,\Rightarrow x+y=10........(4)\]
Now let one number be x and another number be 10-x.
Here it is given that geometric mean is 4, so from above information and equation (2) we get,
 \[\Rightarrow \sqrt{x(10-x)}=4......(5)\]
Squaring both sides of equation (5) we get,
 \[\Rightarrow x(10-x)=16......(6)\]
Rearranging equation (6) we get,
 \[\Rightarrow {{x}^{2}}-10x+16=0......(7)\]
Now factorizing equation (7) we get,
 \[\begin{align}
  & \Rightarrow {{x}^{2}}-8x-2x+16=0 \\
 & \Rightarrow x(x-8)-2(x-8)=0 \\
 & \Rightarrow (x-2)\,(x-8)=0 \\
 & \Rightarrow x=2\,,8 \\
\end{align}\]
We got the two numbers as 2, 8.
Now the harmonic mean between two numbers is defined as \[=\dfrac{2ab}{a+b}\].
Using this formula, we get harmonic mean between 2 and 8 as,
 \[\Rightarrow \dfrac{2\times 2\times 8}{2+8}=\dfrac{32}{10}=\dfrac{16}{5}\]
Hence the harmonic mean is \[\dfrac{16}{5}\].
Now coming to second part of the question,
We know that the terms of H.P are reciprocal of terms of A.P.
Let a be the first term and d be the common difference of corresponding A.P.
The nth term of arithmetic progression is \[a+(n-1)d\] whereas the nth term of geometric progression is \[\,\dfrac{1}{a+(n-1)d}\].
Third term and fifth term is given and we need to find tenth term. By converting these terms into terms of A.P from the above definition we get,
 \[\begin{align}
  & {{T}_{3}}=\dfrac{1}{a+(3-1)d}=1 \\
 & \,\Rightarrow \dfrac{1}{a+2d}=1 \\
 & \,\Rightarrow a+2d=1.......(1) \\
\end{align}\]
 \[\begin{align}
  & {{T}_{5}}=\dfrac{1}{a+(5-1)d}=-5 \\
 & \,\Rightarrow \dfrac{1}{a+4d}=\dfrac{-1}{5} \\
 & \,\Rightarrow a+4d=-5.......(2) \\
\end{align}\]
Solving equation (1) and equation (2) simultaneously we get \[a=7\] and \[\,d=-3\].
Using the value of a and d we get,
 \[\begin{align}
  & {{T}_{10}}=\dfrac{1}{a+(10-1)d} \\
 & \,\Rightarrow \dfrac{1}{a+9d}=\dfrac{1}{7+9\times (-3)}=-\dfrac{1}{20} \\
\end{align}\]
So the tenth term of H.P is \[-\dfrac{1}{20}\].

Note: Knowing formula of arithmetic mean, geometric mean and harmonic mean is important. We may get confused by harmonic progression but here we just have to remember that the terms in H.P are reciprocal of the terms in A.P. Knowing the nth term formula of A.P, G.P is also important because this will help reduce the time taken to solve.