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Question: The first term of an arithmetic sequence is equal to 200 and the common difference is equal to \[ - ...

The first term of an arithmetic sequence is equal to 200 and the common difference is equal to 10 - 10. Find the value of the 20th term.

Explanation

Solution

Here, we will find the common difference and then use the formula of nnth term of the arithmetic progression A.P., that is, an=a+(n1)d{a_n} = a + \left( {n - 1} \right)d, where aa is the first term and dd is the common difference. Apply this formula, and then substitute the value of aa,dd and nn in the obtained equation to find the value of the required term.

Complete step by step answer:

We are given that the first term of an arithmetic sequence is 200 and the common difference is 10 - 10.

We know that the arithmetic progression is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value.

We will now find the value of first term aa and the common differencedd, we get
a=200a = 200
d=10d = - 10

We will use the formula of nnth term of the arithmetic progression A.P., that is, an=a+(n1)d{a_n} = a + \left( {n - 1} \right)d, where aa is the first term and dd is the common difference.

We know that n=20n = 20.
Substituting these values of nn, aa and dd in the above formula for the sum of the arithmetic progression, we get

a20=200+(201)(10) a20=200+19(10) a20=200190 a20=10  \Rightarrow {a_{20}} = 200 + \left( {20 - 1} \right)\left( { - 10} \right) \\\ \Rightarrow {a_{20}} = 200 + 19\left( { - 10} \right) \\\ \Rightarrow {a_{20}} = 200 - 190 \\\ \Rightarrow {a_{20}} = 10 \\\

Thus, the 20th term of the given AP is 10.

Note: In solving these types of questions, you should be familiar with the formula of sum of the arithmetic progression and their sums. Some students use the formula to find the sum, S=n2(a+l)S = \dfrac{n}{2}\left( {a + l} \right), where ll is the last term, but we do not have the to find the value of an{a_n} so it will be wrong. We can also find the value of nnth term by find the value of SnSn1{S_n} - {S_{n - 1}}, where Sn=n2(2a+(n1)d){S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right), where aa is the first term and dd is the common difference. But this is a longer method, which takes time, so we will use the above method. One should know the an{a_n} is the nnth term in the geometric progression.