Question

Question:-


The 17th term of an A.P. is 14 more than its 10th term. Find the common difference.​

Answer 1

Given:-

• 17th term of an AP is 14 more than 10th term.

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To Find:-

• common difference?

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Solution:-

$\underline{\bigstar\:\boldsymbol{According\:to\:the\: Question\::}}$

$:\implies\sf a_{17} = a_{10} + 14$

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$:\implies\sf a + (17-1) d = a + (10-1)d +14$

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$:\implies\sf a+ 16d = (a + 9d) + 14$

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$:\implies\sf a+ 16d - (a+9d) = 14$

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$:\implies\sf a+ 16d - a - 9d = 14$

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$:\implies\sf 7d = 14$

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$:\implies\sf d = \cancel{ \dfrac{14}{7}}$

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$:\implies{\boxed{\sf{\purple{d = 2}}}}\;\bigstar$

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$\therefore\;{\underline{\sf{Hence,\;Common\; difference\;of\;an\;AP\:is\; \bf{2}.}}}$

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Answer 2

Given:-

• 17th term of an AP is 14 more than 10th term.

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To Find:-

• common difference

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Solution:-

$\begin{gathered}\underline{\bigstar\:\boldsymbol{According\:to\:the\: Question\::}}\\\end{gathered}$

$\begin{gathered}:\mapsto\sf a_{17} = a_{10} + 14\\\end{gathered}$

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$\begin{gathered}:\mapsto\sf a + (17-1) d = a + (10-1)d +14\\\end{gathered}$

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$\begin{gathered}:\mapsto\sf a+ 16d = (a + 9d) + 14\\\end{gathered}$

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$\begin{gathered}:\mapsto\sf a+ 16d - (a+9d) = 14\\\end{gathered}$

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$\begin{gathered}:\mapsto\sf a+ 16d - a - 9d = 14\\\end{gathered}$

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$\begin{gathered}:\mapsto\sf 7d = 14\\\end{gathered}$

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$\begin{gathered}:\mapsto\sf d = \cancel{ \dfrac{14}{7}}\\\end{gathered}$

$\begin{gathered}:\mapsto{\boxed{\sf{\pink{d = 2}}}}\;\bigstar\\\end{gathered}$

$\therefore\;{\underline{\sf{Hence,\;Common\; difference\;of\;an\;AP\:is\; \bf{2}.}}}$

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