
Find the $ {{20}^{th}} $ term of $ -5,\dfrac{-5}{2},0,\dfrac{5}{2},.... $ .
Answer
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Hint: First, we have to check whether the series given is arithmetic progression (AP) or not. This can be checked by taking the difference between 2 consecutive numbers that should be the same for all. So, on taking the difference between the first 2 numbers, we get difference d as $ d=\dfrac{-5}{2}-\left( -5 \right)=\dfrac{5}{2} $ . Now, taking for next 2 numbers we get as $ d=\dfrac{5}{2}-0=\dfrac{5}{2} $ . So, this is in AP. Now, further we will use the formula for finding $ {{n}^{th}} $ term in series using the formula $ {{T}_{n}}=a+\left( n-1 \right)d $ where a is first number of series, d is common difference between two consecutive number, n is the number of term which we want to find which is 20 here. Thus, on solving we will get the answer.
Complete step-by-step answer:
Here, we will use the formula of Arithmetic progression (AP) for finding $ {{n}^{th}} $ term in series i.e. $ {{T}_{n}}=a+\left( n-1 \right)d $ where a is first number of series, d is common difference between two consecutive number i.e. $ d=\dfrac{-5}{2}-\left( -5 \right)=\dfrac{5}{2} $ , n is the number of terms which we want to find which is 20 here.
So, on substituting the values, we get as
$ {{T}_{20}}=-5+\left( 20-1 \right)\cdot \dfrac{5}{2} $
On further solving, we get as
$ {{T}_{20}}=-5+\dfrac{19\times 5}{2} $
$ {{T}_{20}}=-5+\dfrac{95}{2} $
We will take LCM and on solving, we will get as
$ {{T}_{20}}=\dfrac{-5\times 2+95}{2} $
$ {{T}_{20}}=\dfrac{-10+95}{2}=\dfrac{85}{2} $
Thus, $ {{20}^{th}} $ term of $ -5,\dfrac{-5}{2},0,\dfrac{5}{2},.... $ is $ \dfrac{85}{2} $ .
Note: Be careful in finding common differences d. Sometimes students make mistakes in taking differences as $ d=-5-\left( \dfrac{-5}{2} \right)=\dfrac{-5}{2} $ . And now, on substituting the values in the equation, we get as $ {{T}_{20}}=-5-\left( 20-1 \right)\cdot \dfrac{5}{2} $ . On solving, we get answer to be $ {{T}_{20}}=-5-\dfrac{95}{2}=\dfrac{-105}{2} $ which will be wrong. So, do not make this mistake and be clear with common difference concepts.
Complete step-by-step answer:
Here, we will use the formula of Arithmetic progression (AP) for finding $ {{n}^{th}} $ term in series i.e. $ {{T}_{n}}=a+\left( n-1 \right)d $ where a is first number of series, d is common difference between two consecutive number i.e. $ d=\dfrac{-5}{2}-\left( -5 \right)=\dfrac{5}{2} $ , n is the number of terms which we want to find which is 20 here.
So, on substituting the values, we get as
$ {{T}_{20}}=-5+\left( 20-1 \right)\cdot \dfrac{5}{2} $
On further solving, we get as
$ {{T}_{20}}=-5+\dfrac{19\times 5}{2} $
$ {{T}_{20}}=-5+\dfrac{95}{2} $
We will take LCM and on solving, we will get as
$ {{T}_{20}}=\dfrac{-5\times 2+95}{2} $
$ {{T}_{20}}=\dfrac{-10+95}{2}=\dfrac{85}{2} $
Thus, $ {{20}^{th}} $ term of $ -5,\dfrac{-5}{2},0,\dfrac{5}{2},.... $ is $ \dfrac{85}{2} $ .
Note: Be careful in finding common differences d. Sometimes students make mistakes in taking differences as $ d=-5-\left( \dfrac{-5}{2} \right)=\dfrac{-5}{2} $ . And now, on substituting the values in the equation, we get as $ {{T}_{20}}=-5-\left( 20-1 \right)\cdot \dfrac{5}{2} $ . On solving, we get answer to be $ {{T}_{20}}=-5-\dfrac{95}{2}=\dfrac{-105}{2} $ which will be wrong. So, do not make this mistake and be clear with common difference concepts.
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