find the h.p. whose 6th term is 1 and 10th term is 5. find 15th term of the series also.

so term(6) = 1/(a+5d) = 1

a+5d = 1 **

term(10) = 1/(a+9d) = 5
5a + 45d = 1 ***

** times 5 ---> 5a + 25d = 5
subtract from ***
20d = -4
d = -1/5
in **
a - 1 = 1
a = 2
so the HP is 1/2 , 5/9 5/8, 27/20, ...
term(15) = 1/(a+14d)
= 1/(2 - 14/5)
= -5/4

To find the common difference and the first term (a) of an arithmetic series, we can analyze the given information.

Given:
6th term (a₆) = 1
10th term (a₁₀) = 5

Step 1: Find the common difference (d)
To find the common difference, we use the formula for the general term of an arithmetic series:

aₙ = a + (n-1)d

Where:
aₙ = nth term
a = first term
d = common difference
n = term number

We have two equations based on the given information:
a₆ = a + (6-1)d = 1 ...(Equation 1)
a₁₀ = a + (10-1)d = 5 ...(Equation 2)

Step 2: Solve for a and d
To solve for a and d, we will solve the system of equations formed by Equation 1 and Equation 2:

From Equation 1, we have: a + 5d = 1
From Equation 2, we have: a + 9d = 5

Subtracting the two equations, we get:
4d = 4
Dividing both sides by 4, we find: d = 1

Plugging d = 1 into Equation 1, we get:
a + 5 = 1
Subtracting 5 from both sides, we find: a = -4

So, the first term (a) is -4, and the common difference (d) is 1.

Step 3: Find the 15th term (a₁₅)
To find the 15th term, we use the formula for the general term of an arithmetic series:

aₙ = a + (n-1)d

Plugging in the values:
a = -4
d = 1
n = 15

We have:
a₁₅ = -4 + (15-1) * 1
= -4 + 14
= 10

Therefore, the 15th term of the series is 10.

To find the common ratio of the harmonic progression (H.P.), we need to first find the value of the first term.

Let's assume the first term of the H.P. is 'a' and the common ratio is 'd'.

Given:
6th term = 1
10th term = 5

The formula for the nth term of an H.P. is:
\(A_n = \frac{1}{a + (n-1)d}\)

Using this formula, we can set up two equations using the given information:

For the 6th term (n = 6):
\(A_6 = \frac{1}{a + 5d} = 1\) ----(1)

For the 10th term (n = 10):
\(A_{10} = \frac{1}{a + 9d} = 5\) ----(2)

To find 'a' and 'd', we need to solve these two equations simultaneously.

First, let's solve equation (1) for 'd':
\(a + 5d = 1\)
\(5d = 1 - a\)
\(d = \frac{1 - a}{5}\) ----(3)

Now, substitute equation (3) into equation (2):
\(A_{10} = \frac{1}{a + 9d}\)
\(5 = \frac{1}{a + 9\left(\frac{1 - a}{5}\right)}\)
\(5 = \frac{5}{a + 9 - 9a}\)
\(5(a + 9 - 9a) = 5\)
\(5a + 45 - 45a = 5\)
\(45 - 40a = 5\)
\(40a = 40\)
\(a = 1\) ----(4)

Now that we have the value of 'a', we can substitute it back into equation (3) to find the value of 'd':
\(d = \frac{1 - a}{5}\)
\(d = \frac{1 - 1}{5}\)
\(d = 0\) ----(5)

So, the first term 'a' is 1 and the common ratio 'd' is 0.

Now let's find the 15th term of the series.

The formula for the nth term of an H.P. is:
\(A_n = \frac{1}{a + (n-1)d}\)

For the 15th term (n = 15):
\(A_{15} = \frac{1}{1 + (15-1) \times 0}\)
\(A_{15} = \frac{1}{1}\)
\(A_{15} = 1\)

Therefore, the 15th term of the series is 1.