Question

How do you find the 10th term of an arithmetic series?
The 3rd term of an arithmetic series A is 19.
The sum of the first 10 terms of A is 290
Find the 10th term of A.
Use \[{{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\]

Answer 1

Hint: To solve these problems, we should know the following properties of the arithmetic series. The nth term of an arithmetic series is \[a+\left( n-1 \right)d\]. Here, a is the first term of series, d is the common difference, and n is the position of the term. The sum of first n terms of an arithmetic series is evaluated as \[{{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\]. For this question, we will form two linear equations in two variables. On solving this set of equations, we will get the first term and the common difference of the arithmetic series.

Complete step by step solution:
Let the first term of the arithmetic series A is a, and the common difference is d. Using this, we can write an expression for the nth term of the series as \[a+\left( n-1 \right)d\]. Also, we can write the expression for the sum of first n terms of an arithmetic series is as \[{{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\].
We are given that the 3rd term of the arithmetic series is 19. Hence, we can write that \[a+(3-1)d=19\]. simplifying this, we get \[a+2d=19\].
Similarly, we are also given that the sum of the first 10 terms is 290. Hence, we can use the summation expression, \[\dfrac{10}{2}\left( 2a+\left( 10-1 \right)d \right)=290\]. Simplifying this, we get \[2a+9d=58\].
Now, we have two equations \[a+2d=19\] and \[2a+9d=58\]. We can solve these two equations to find the values of a and d. on solving, we get \[d=4\And a=11\].
We need to find the 10th term of the series. Using the expression for the nth term, we get the 10th term as \[11+\left( 10-1 \right)4=11+36=47\].

Note: To solve the given question, one must know the properties and expressions for nth term and sum of n terms of different types of series. For example, in the case of geometric series where the first term is 1 and the common ratio is r. The nth term is represented as \[a{{r}^{n-1}}\] and the sum of first n terms is \[\dfrac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}\].