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Arithmetic Progressions
You can see apart of an auditorium. The first row has 25 seats. the second row has 7 more seats than a first row. The number of seats in each succeeding row increases by fixed number 7. How many seats will be there in the 10th row. How many rows will the auditorium need so that the capacity is 1110 people. This is an example of Arithmetic Progression. You will be able to solve the problem at the end of this module.
Derivation of the nth term
Derivation of sum of first n terms
Let T1, T2, T3 and so on upto Tn be the terms of an arithmetic progression.
Let d be the common difference between the successive terms
T2 minus T1 is equal to T3 minus T2 is equal to T4 minus T3 and So on i.e., equal to Tn minus Tn minus 1 i.e., equal to d
The first term T1 is a and it can be written as a plus 1 minus 1 into d.
Follow the same pattern we can write the second term as a plus 2 minus 1 into d.
Now write T3 and T4 using the same pattern key in the answers
Follow the same pattern we get the nth term Tn is equal to the n minus 1th term i.e., Tn minus 1 plus d i.e., equal to a plus n minus 1 into d.
The general form for the nth term of an arithmetic progression is Tn is equal to a plus n minus 1 into d.
In the example of the auditorium the number of seats in the first row i.e., a is equal to 25 the common difference d is equal to 7. Find the number of seats in the 10 th row. we use the formula for the n term and substitute the known values. key in the values. key in the answer.
Find there are 48 seats in the 10 th row
Let a be the first term of an arithmetic progression of n terms and let the common difference between the successive terms be d write the term of the arithmetic progression key in the values.
let us denote the sum of first n terms of an arithmetic progression by Sn we get Sn is equal to a plus a plus d a plus 2d a plus 3d plus so on till a plus n minus 1 times d. That equals n times a plus d plus 2d plus 3d and so on upto n minus 1 times d. i.e., n times a plus sigma n minus 1 times d.
We know that sigma n i.e., the sum of first n natural numbers is equal to n into n plus 1 by 2. In the same way sigma n minus 1 i.e., the sum of first n minus 1 natural numbers is equal to n minus 1into n minus plus 1 by 2.
We substitute this value of sigma n minus 1 in the right hand side of the equation. we simplify the terms we get the formula Sn equal to n by 2 times 2a plus n minus 1 into d. We use this formula when we know the values of a, n and d. We can rearrange the terms we get the formula Sn is equal ton by 2 times a plus l where l is the last term of the arithmetic progression. We use this formula when we know the Values a, n and l. In the example of the auditorium the number of the seats in the first row i.e. a is equal to 25 and the common difference d is equal to 7. How many rows will the auditorium need so that the capacity is 1110 people. We use the formula for the sum of nth terms and substitute the known values. Key in the values.
We reject the negative value we find that 15 rows are needed to sit 1110 people in the auditorium.
Watch Video
Arithmetic Progressions
You can see apart of an auditorium. The first row has 25 seats. the second row has 7 more seats than a first row. The number of seats in each succeeding row increases by fixed number 7. How many seats will be there in the 10th row. How many rows will the auditorium need so that the capacity is 1110 people. This is an example of Arithmetic Progression. You will be able to solve the problem at the end of this module.
Derivation of the nth term
Derivation of sum of first n terms
Let T1, T2, T3 and so on upto Tn be the terms of an arithmetic progression.
Let d be the common difference between the successive terms
T2 minus T1 is equal to T3 minus T2 is equal to T4 minus T3 and So on i.e., equal to Tn minus Tn minus 1 i.e., equal to d
The first term T1 is a and it can be written as a plus 1 minus 1 into d.
Follow the same pattern we can write the second term as a plus 2 minus 1 into d.
Now write T3 and T4 using the same pattern key in the answers
Follow the same pattern we get the nth term Tn is equal to the n minus 1th term i.e., Tn minus 1 plus d i.e., equal to a plus n minus 1 into d.
The general form for the nth term of an arithmetic progression is Tn is equal to a plus n minus 1 into d.
In the example of the auditorium the number of seats in the first row i.e., a is equal to 25 the common difference d is equal to 7. Find the number of seats in the 10 th row. we use the formula for the n term and substitute the known values. key in the values. key in the answer.
Find there are 48 seats in the 10 th row
Let a be the first term of an arithmetic progression of n terms and let the common difference between the successive terms be d write the term of the arithmetic progression key in the values.
let us denote the sum of first n terms of an arithmetic progression by Sn we get Sn is equal to a plus a plus d a plus 2d a plus 3d plus so on till a plus n minus 1 times d. That equals n times a plus d plus 2d plus 3d and so on upto n minus 1 times d. i.e., n times a plus sigma n minus 1 times d.
We know that sigma n i.e., the sum of first n natural numbers is equal to n into n plus 1 by 2. In the same way sigma n minus 1 i.e., the sum of first n minus 1 natural numbers is equal to n minus 1into n minus plus 1 by 2.
We substitute this value of sigma n minus 1 in the right hand side of the equation. we simplify the terms we get the formula Sn equal to n by 2 times 2a plus n minus 1 into d. We use this formula when we know the values of a, n and d. We can rearrange the terms we get the formula Sn is equal ton by 2 times a plus l where l is the last term of the arithmetic progression. We use this formula when we know the Values a, n and l. In the example of the auditorium the number of the seats in the first row i.e. a is equal to 25 and the common difference d is equal to 7. How many rows will the auditorium need so that the capacity is 1110 people. We use the formula for the sum of nth terms and substitute the known values. Key in the values.
We reject the negative value we find that 15 rows are needed to sit 1110 people in the auditorium.
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