Answer

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**Hint:**A sequence is a list of items/objects which have been arranged in a sequential way.

A series can be highly generalized as the sum of all the terms in a sequence however, there has to be a definite relationship between all the terms of the sequence.

Arithmetic sequence:

A sequence in which every term is created by adding or subtracting a definite number to the preceding number is an arithmetic sequence.

Sequence - \[a,\,\,a + d\,,\,\,a + 2d.....\]

- \[a\left( {n - 1} \right)d.....\]

General term (nth term) - \[{a_n} = a + (n - 1)d\]

**Complete step by step solution:**

Let, no. of pens in the first year be ‘a’.

And uniform increase be ‘d’

From the given statement:

We have,

\[a{}_5 = a + \left( {5 - 1} \right)d\]

\[a + 4d = 16000\] ______ (1).

By using the formula \[\left( {a + \left( {n - 1} \right)d} \right)\]

Where n \[ = 5\]year

And in \[{18^{th}}\]year factory manufactured \[20500\]pens.

Now \[{a_{18}} = a + \left( {18 - 1} \right)d\]

\[a + 17d = 20500\] _______ (2).

By subtracting the equation (1) from equation (2).

\[a + 17d - a - 4d = 20500 - 16000\]

\[13d = 4500\]

\[d = \dfrac{{4500}}{{13}}\]

\[d = 300\].

By putting the volume of ‘d’ in equation (1)

We get,

\[a + 4 \times 300 = 16000\]

\[a + 1200 = 16000\]

\[a = 16000 - 1200\]

\[a = 14800\].

(ii) The total production in \[10\]years

For \[10\]have:

\[a + 9d\] ________ (3)

By putting the value of ‘a’

\[14800 + 9 \times 300\]

\[ = 17500\].

In \[10\]years we having two leap years so that total production is \[10\]is equal to:

\[ = \dfrac{7}{2} \times \left( {14800 + 17500} \right)\]

\[ = 7 \times 16150\]

\[ = 113050\].

Hence, the number of pens manufactured in the first year \[14800\] pens.

And the total production in \[10\]years \[113050\]pens.

**Note:**In a question if it is given that the quantity is uniformly increasing, this means to say that the sequence is an Arithmetic Progression(A.P)

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