
If ${{A}_{1}},{{A}_{2}},.........{{A}_{n}}$ are n arithmetic means between a and b. Find the common difference between the terms.
Answer
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Hint: As it is given that ${{A}_{1}},{{A}_{2}},.........{{A}_{n}}$ are n arithmetic means between a and b, we can deduce that the sequence $a,{{A}_{1}},{{A}_{2}},.........{{A}_{n}},b$ is an A.P with n+2 terms. b is the last term of the A.P., so use the formula ${{T}_{r}}=a+\left( r-1 \right)d$ and substitute the known variables 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 ${{T}_{r}}$, and sum till r terms is denoted by ${{S}_{r}}$ .
${{T}_{r}}=a+\left( r-1 \right)d$
${{S}_{r}}=\dfrac{r}{2}\left( 2a+\left( r-1 \right)d \right)=\dfrac{r}{2}\left( a+l \right)$
Now moving to the situation given in the question. As it is given that ${{A}_{1}},{{A}_{2}},.........{{A}_{n}}$ are n arithmetic means between a and b, we can deduce that the sequence $a,{{A}_{1}},{{A}_{2}},.........{{A}_{n}},b$ is an A.P with n+2 terms. Also, looking at the sequence we can say that the first term of the sequence is a and there is a total of (n+2) terms. Also, the last term of the A.P. is b.
So, if we use the formula of general term of a A.P., we get
${{T}_{n+2}}=a+\left( n+2-1 \right)d$
$\Rightarrow b=a+\left( n+2-1 \right)d$
$\Rightarrow b-a=\left( n+1 \right)d$
$\Rightarrow d=\dfrac{b-a}{n+1}$
Hence, we can conclude that the common difference is $\dfrac{b-a}{n+1}$ .
Note: Always, while dealing with an arithmetic progression, try to find the first term and the common difference of the sequence, as once you have these quantities, you can easily solve the questions related to the given arithmetic progressions. Also, remember that the answer must be reported in the simplest form and in terms variables given in the question. For instance: the common difference for the above question can be ${{A}_{2}}-{{A}_{1}}$ as well, but we needed the answer to be in the terms of a and b.
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 ${{T}_{r}}$, and sum till r terms is denoted by ${{S}_{r}}$ .
${{T}_{r}}=a+\left( r-1 \right)d$
${{S}_{r}}=\dfrac{r}{2}\left( 2a+\left( r-1 \right)d \right)=\dfrac{r}{2}\left( a+l \right)$
Now moving to the situation given in the question. As it is given that ${{A}_{1}},{{A}_{2}},.........{{A}_{n}}$ are n arithmetic means between a and b, we can deduce that the sequence $a,{{A}_{1}},{{A}_{2}},.........{{A}_{n}},b$ is an A.P with n+2 terms. Also, looking at the sequence we can say that the first term of the sequence is a and there is a total of (n+2) terms. Also, the last term of the A.P. is b.
So, if we use the formula of general term of a A.P., we get
${{T}_{n+2}}=a+\left( n+2-1 \right)d$
$\Rightarrow b=a+\left( n+2-1 \right)d$
$\Rightarrow b-a=\left( n+1 \right)d$
$\Rightarrow d=\dfrac{b-a}{n+1}$
Hence, we can conclude that the common difference is $\dfrac{b-a}{n+1}$ .
Note: Always, while dealing with an arithmetic progression, try to find the first term and the common difference of the sequence, as once you have these quantities, you can easily solve the questions related to the given arithmetic progressions. Also, remember that the answer must be reported in the simplest form and in terms variables given in the question. For instance: the common difference for the above question can be ${{A}_{2}}-{{A}_{1}}$ as well, but we needed the answer to be in the terms of a and b.
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