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Question: The first term of an AP is 5, the common difference is 3 and the last term is 80. Find the number of...

The first term of an AP is 5, the common difference is 3 and the last term is 80. Find the number of terms.

Explanation

Solution

We solve this problem by using the nth{{n}^{th}} term of an AP.
If ′a′'a' is the first term and ′d′'d' is the common difference of an AP then the nth{{n}^{th}} term of the AP is given as,
Tn=a+(n−1)d{{T}_{n}}=a+\left( n-1 \right)d
By using this formula to the given data we find the required number of terms by using the condition that the value of ′n′'n' for the last term will be the number of terms of given AP.

Complete step by step solution:
We are given that the first term of the AP as 5
Let us assume that the first term of the AP as,
⇒a=5\Rightarrow a=5
We are also given the common difference of the AP as 3.
Let us assume that the given common difference of the AP as,
⇒d=3\Rightarrow d=3
We are asked to find the total number of terms in the AP if the last term is 80.
Let us assume that there are ′n′'n' number of terms in the AP and assume that last term as nth{{n}^{th}} term of AP that is,
⇒Tn=80\Rightarrow {{T}_{n}}=80
Now, let us use the formula of nth{{n}^{th}} term of the AP.
We know that the nth{{n}^{th}} term of the AP is given as,
Tn=a+(n−1)d{{T}_{n}}=a+\left( n-1 \right)d
Where, ′a′'a' is the first term and ′d′'d' is the common difference of an AP.
By using this formula for the given data then we get,
⇒80=5+(n−1)(3) ⇒3n−3=75 ⇒3n=78 \begin{aligned} & \Rightarrow 80=5+\left( n-1 \right)\left( 3 \right) \\\ & \Rightarrow 3n-3=75 \\\ & \Rightarrow 3n=78 \\\ \end{aligned}
By dividing the both sides of the equation with 3 we get,
⇒3n3=783 ⇒n=26 \begin{aligned} & \Rightarrow \dfrac{3n}{3}=\dfrac{78}{3} \\\ & \Rightarrow n=26 \\\ \end{aligned}

Therefore, we can conclude that there are a total of 26 terms in the given AP.

Note: Students may make a mistake by adding ‘1’ to the value of ′n′'n' because they misunderstand that the first term is not counted. In the formula of Tn{{T}_{n}} the value ′n′'n' represents the number of the term.
If n=1n=1 then it says that T1{{T}_{1}} which is the first term of AP. So, if the last term has n=26n=26 then we can directly say that there are 26 terms. There will be no need to add ‘1’.