If the $ {{n}^{th}} $ terms of the two APs: 9, 7, 5,….. and 24,21,18,….. are the same, find the value of n.
Answer
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Hint: If you observe the AP 9, 7, 5,….., you will find that the first term is 9 and the common difference is -2. Similarly, for the other AP 24,21,18,….., the first term is 24 and the common difference is -3. So, use the formula of general term of an AP, i.e., $ {{T}_{n}}=a+\left( n-1 \right)d $ for both the APs separately to get the $ {{n}^{th}} $ terms and equate them to get the answer.
Complete step-by-step answer:
Before starting with the solution, let us discuss what an A.P. is. A.P. stands for arithmetic progression and is defined as a sequence of numbers for which the difference of two consecutive terms is constant. The general term of an arithmetic progression is denoted by $ {{T}_{n}} $ , and sum till n terms is denoted by $ {{S}_{n}} $ .
$ {{T}_{n}}=a+\left( n-1 \right)d $
$ {{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right) $
In the above formulas ‘a’ is the first term and ‘d’ is the common difference.
Now, let us start the solution to the above question by finding the $ {{n}^{th}} $ term of the AP 9, 7, 5,…… The first term of the AP is 9 and the common difference is -2. So, using the formula of $ {{n}^{th}} $ of the AP, we get
$ {{T}_{n}}=a+\left( n-1 \right)d=9+\left( n-1 \right)\left( -2 \right)=9-2n+2=11-2n $
Now, let us move to the AP 24,21,18,….. For this AP, a=24 and d=-3. If we again use the formula of $ {{n}^{th}} $ of the AP, we get
$ {{T}_{n}}=a+\left( n-1 \right)d=24+\left( n-1 \right)\left( -3 \right)=24-3n+3=27-3n $
Now, as it is given that the $ {{n}^{th}} $ terms of the two APs are equal, we will equate the two results.
$ 11-2n=27-3n $
$ \Rightarrow -2n+3n=27-11 $
$ \Rightarrow n=16 $
Hence, the value of n satisfying the condition given in the question is 16.
Note: In a question related to an arithmetic progression, the most important thing is to find the common difference and the first term. You also need to learn the basic properties of all the standard sequences like the arithmetic progression and geometric progressions. Also, be careful don’t put d to be 2 and 3 while finding the $ {{n}^{th}} $ terms of the two APs instead of -2 and -3, respectively.
Complete step-by-step answer:
Before starting with the solution, let us discuss what an A.P. is. A.P. stands for arithmetic progression and is defined as a sequence of numbers for which the difference of two consecutive terms is constant. The general term of an arithmetic progression is denoted by $ {{T}_{n}} $ , and sum till n terms is denoted by $ {{S}_{n}} $ .
$ {{T}_{n}}=a+\left( n-1 \right)d $
$ {{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right) $
In the above formulas ‘a’ is the first term and ‘d’ is the common difference.
Now, let us start the solution to the above question by finding the $ {{n}^{th}} $ term of the AP 9, 7, 5,…… The first term of the AP is 9 and the common difference is -2. So, using the formula of $ {{n}^{th}} $ of the AP, we get
$ {{T}_{n}}=a+\left( n-1 \right)d=9+\left( n-1 \right)\left( -2 \right)=9-2n+2=11-2n $
Now, let us move to the AP 24,21,18,….. For this AP, a=24 and d=-3. If we again use the formula of $ {{n}^{th}} $ of the AP, we get
$ {{T}_{n}}=a+\left( n-1 \right)d=24+\left( n-1 \right)\left( -3 \right)=24-3n+3=27-3n $
Now, as it is given that the $ {{n}^{th}} $ terms of the two APs are equal, we will equate the two results.
$ 11-2n=27-3n $
$ \Rightarrow -2n+3n=27-11 $
$ \Rightarrow n=16 $
Hence, the value of n satisfying the condition given in the question is 16.
Note: In a question related to an arithmetic progression, the most important thing is to find the common difference and the first term. You also need to learn the basic properties of all the standard sequences like the arithmetic progression and geometric progressions. Also, be careful don’t put d to be 2 and 3 while finding the $ {{n}^{th}} $ terms of the two APs instead of -2 and -3, respectively.
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