# If for an arithmetic progression, d=11, what is ${{t}_{17}}-{{t}_{15}}?$.

Answer

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Hint: In an arithmetic progression sequence, the nth term ${{t}_{n}}$ is defined by ${{t}_{n}}=a+(n-1)d$ where a is the first term and d is the common difference of the arithmetic progression.

Complete step-by-step answer:

Let us consider the sequence of numbers as a, a+d, a+2d, a+3d ………. a+(n-1)d, in which successive numbers differ by a constant d. This constant is called the common difference (d) and a is the first term. This type of sequence is called the arithmetic progression.

From the above sequence, we can define the general term ${{t}_{n}}$ as ${{t}_{n}}=a+(n-1)d$.

In an arithmetic progression, ${{t}_{n}}$ means the nth term in the series.

In the same way, ${{t}_{17}}$ is the 17th term of an arithmetic progression and ${{t}_{15}}$ is the 15th term of an arithmetic progression.

Now we will find the term ${{t}_{17}}$ of the arithmetic progression by using the formula for the nth term of the arithmetic progression. So, we will get,

${{t}_{17}}=a+(17-1)d$

$\Rightarrow a+16d$, let this be equation (1).

Now we will express ${{t}_{15}}$ of the arithmetic progression by using the formula for the nth term of the arithmetic progression. So, we will get,

${{t}_{15}}=a+(15-1)d$

$\Rightarrow a+14d$ , let this be equation (2).

According to the question, we are asked to find ${{t}_{17}}-{{t}_{15}}$.

So, we will subtract equation (2) from equation (1), and we will get,

$\Rightarrow {{t}_{17}}-{{t}_{15}}=a+16d-(a+14d)$

$\Rightarrow 2d$ let it be equation (3).

According to the question, the common difference is given as d=11.

Now we will substitute d=11 in equation (3) and we will get,

$\begin{align}

& {{t}_{17}}-{{t}_{15}}=2\times d \\

& {{t}_{17}}-{{t}_{15}}=2\times 11 \\

& {{t}_{17}}-{{t}_{15}}=22 \\

\end{align}$

Hence, we have found that ${{t}_{17}}-{{t}_{15}}$=22.

Note: The common difference of an arithmetic progression can be negative as well as positive. It is necessary to remember the formula to find the general term of an arithmetic progression. The common mistake that a student can make is considering the formula as ${{t}_{n}}=a+(n+1)d$ instead of using the formula ${{t}_{n}}=a+(n-1)d$. This can lead to the wrong answer.

Complete step-by-step answer:

Let us consider the sequence of numbers as a, a+d, a+2d, a+3d ………. a+(n-1)d, in which successive numbers differ by a constant d. This constant is called the common difference (d) and a is the first term. This type of sequence is called the arithmetic progression.

From the above sequence, we can define the general term ${{t}_{n}}$ as ${{t}_{n}}=a+(n-1)d$.

In an arithmetic progression, ${{t}_{n}}$ means the nth term in the series.

In the same way, ${{t}_{17}}$ is the 17th term of an arithmetic progression and ${{t}_{15}}$ is the 15th term of an arithmetic progression.

Now we will find the term ${{t}_{17}}$ of the arithmetic progression by using the formula for the nth term of the arithmetic progression. So, we will get,

${{t}_{17}}=a+(17-1)d$

$\Rightarrow a+16d$, let this be equation (1).

Now we will express ${{t}_{15}}$ of the arithmetic progression by using the formula for the nth term of the arithmetic progression. So, we will get,

${{t}_{15}}=a+(15-1)d$

$\Rightarrow a+14d$ , let this be equation (2).

According to the question, we are asked to find ${{t}_{17}}-{{t}_{15}}$.

So, we will subtract equation (2) from equation (1), and we will get,

$\Rightarrow {{t}_{17}}-{{t}_{15}}=a+16d-(a+14d)$

$\Rightarrow 2d$ let it be equation (3).

According to the question, the common difference is given as d=11.

Now we will substitute d=11 in equation (3) and we will get,

$\begin{align}

& {{t}_{17}}-{{t}_{15}}=2\times d \\

& {{t}_{17}}-{{t}_{15}}=2\times 11 \\

& {{t}_{17}}-{{t}_{15}}=22 \\

\end{align}$

Hence, we have found that ${{t}_{17}}-{{t}_{15}}$=22.

Note: The common difference of an arithmetic progression can be negative as well as positive. It is necessary to remember the formula to find the general term of an arithmetic progression. The common mistake that a student can make is considering the formula as ${{t}_{n}}=a+(n+1)d$ instead of using the formula ${{t}_{n}}=a+(n-1)d$. This can lead to the wrong answer.

Last updated date: 25th Sep 2023

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