Question

The 17th term of an A.P. is 7 more than the 10th term. Find the common difference (d).​

Answer 1

Answer:

• d = 1

Step-by-step explanation:

Given:

• The 17th term of an A.P. is 7 more than the 10th term.

To find:

• Common difference

Solution:

In this kind of question, we are going to arrive an equation. From the equation, we will get the value of Common Difference(d), which is our required answer.

$\begin{gathered}\underline{\tt{Formula\:for\:nth\:term:}}\\\\\\:\implies{\boxed{\pink{\sf{a_n = a+(n-1)d}}}}\\\\\\\underline{\tt{Hence,}}\\\\\\:\implies\sf{a_{17} = a+(17-1)d}\\\\\\:\implies\sf{a_{17} = a+16d}\\\\\\\underline{\tt{Similarly,}}\\\\\\:\implies\sf{a_{10} = a+(10-1)d}\\\\\\:\implies\sf{a_{10} = a+9d}\end{gathered}$

$\begin{gathered}\dag{\underline{\textit{\textbf{According\: to\: the\: question:}}}}\\\\\\:\implies\sf{a_{17} = 7+a_{10}}\\\\\\:\implies\sf{a+16d = 7+a+9d}\\\\\\:\implies\sf{\cancel{a}+16d = 7+\cancel{a}+9d}\\\\\\:\implies\sf{16d = 7+9d}\\\\\\:\implies\sf{16d-9d = 7}\\\\\\:\implies\sf{7d = 7}\\\\\\:\implies\sf{d = \dfrac{7}{7}}\\\\\\:\implies\boxed{\red{\bf{d = \frak{1}}}}\checkmark\\\\\\\therefore\underline{\sf{The\:value\:of\:Common\:Difference\:is\:1}}\end{gathered}$

Answer 2

Hey there, here is your AnsweR,

$\\\\ \quad \underline{ \underline{ \large{\purple{ \frak{ \pmb {\pmb {\pmb{Question \: : }}}}}}}}$

• The 17th term of an A.P. is 7 more than the 10th term. Find the common difference (d).

$\\ \quad \underline{ \underline{ \large {\color{purple}{ \frak{ \pmb {\pmb {\pmb{Solution \: : }}}}}}}}$

We know that, for an A.P series,

$\qquad \bull \: \: \tt{a_n = a+(n−1)d}$

Substituting the values :

$: \implies \tt a_{17} = a+(17−1)d \\\ \\ : \implies \tt a_{17} = a +16d$

In the same way,

$: \implies\tt{a_{10} = a+9d}$

$\quad \underline{ \underline{ \large {\color{navy}{ \frak{ \pmb {\pmb {\pmb{Given \:in \: the \: question : }}}}}}}}$

$: \implies\tt a_{17} − a_{10} = 7$

Therefore,

$: \implies\tt(a +16d)−(a+9d) = 7 \\\\\ : \implies\tt{7d = 7}\\\\\ : \implies\tt{d = \cancel\frac{ 7}{7} }\\\\\ \underline { \pmb{\boxed{ : \implies{ \color{purple}\tt{d = 1}}}}}$

.°. The common difference is 1.

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