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After being set up, a pen factory manufactured 16000 pens in the fifth year and 20500 pens in the eighth year. Assuming that production increases uniformly by a fixed number every year find:
(i) The number of pens manufactured in the first year.
(ii) The total production in 10 year.

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Last updated date: 15th Jun 2024
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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)