Question

The 17th term of an A.P. exceeds its 10th term by 7. Find common difference.​

Answer 1

$\underline{\bf{\dag} \:\mathfrak{By \: using \: formula\: :}}$

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$\bf{\dag}\quad\large\boxed{\sf a_n = \bigg(a + (n - 1)d \bigg)}$⠀⠀

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$\bf{Here}\begin{cases}\sf{\:\:\: a = First \: term}\\\sf{\:\:\:n = No. \: of \: terms}\\\sf{\:\:\: d = Common \: difference}\end{cases}$

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Therefore,

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$:\implies\sf a_{17} = a + (17 - 1)d \\\\\\:\implies\sf a_{17} = a + 16d\qquad\quad\bigg\lgroup\bf\: Equation \: (I) \bigg\rgroup$

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Also,

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$:\implies\sf a_{10} = a + (10 - 1)d \\\\\\:\implies\sf a_{10} = a + 9d\qquad\quad\bigg\lgroup\bf\: Equation \: (II) \bigg\rgroup$

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$\large\underline{\textsf{According to Question now :}}$

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• The 17th term of an A.P. exceeds its 10th term by 7.

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$:\implies\sf a_{17} - a_{10} = 7 \\\\\\:\implies\sf \Big(a + 16d \Big) - \Big(a + 9d \Big) = 7 \\\\\\:\implies\sf \cancel{\:a} \: + 16d - \cancel{\:a} \: + 9d = 7 \\\\\\:\implies\sf 16 d - 9d = 7 \\\\\\:\implies\sf 7d = 7 \\\\\\:\implies\sf d = \cancel\dfrac{7}{7} \\\\\\:\implies{\underline{\boxed{\frak{\pink{d_{\:(common\: difference)} = 1}}}}}\:\bigstar$

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$\therefore\:{\underline{\sf{Hence, \ common\: difference \ is \ \bf{1}.}}}$⠀

Answer 2

Step-by-step explanation:

Given:-

• The 17th term of an A.P exceeds its 10th term by 7. Find the common difference.

To find:- We have to find the common difference ?

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$\frak{\underline{\underline{\dag As\ we\ know\ that:-}}}$

$\sf : \implies {a_{n}\ =\ \bigg (a\ +\ 1(n\ -\ 1)\ d \bigg)}$

Here:-

• a, is for the first term.
• n, is for the no. of terms.
• d, is for the common difference.

__________________

$\frak{\underline{\underline{\dag By\ substituting\ the\ values,\ we\ get:-}}}$

$\sf : \implies {a_{17}\ =\ a\ +\ (17\ -\ 1)\ d} \\ \\ \\ \sf : \implies {a_{17}\ =\ a\ +\ 16d.}$

$\sf \therefore {\underline{The\ above\ is\ the\ equation\ (1).}}$

Also,

$\sf : \implies {a_{10}\ =\ a\ +\ (10\ -\ 1)\ d} \\ \\ \\ \sf : \implies {a_{10}\ =\ a\ +\ 9d.}$

$\sf \therefore {\underline{The\ above\ is\ the\ equation\ (2).}}$

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$\frak{\underline{\underline{\dag Now,\ according\ to\ the\ question:-}}}$

Given:-

• The 17th term of an A.P exceeds its 10th term by 7.

$\sf : \implies {a_{17}\ -\ a_{10}\ =\ 7} \\ \\ \\ \sf : \implies {\bigg (a\ +\ 16d \bigg)\ -\ \bigg (a\ +\ 9d \bigg)\ =\ 7} \\ \\ \\ \sf : \implies {a\ +\ 16d\ -\ a\ +\ 9d\ =\ 7} \\ \\ \\ \sf : \implies {16d\ -\ 9d\ =\ 7} \\ \\ \\ \sf : \implies {7d\ =\ 7} \\ \\ \\ \sf : \implies {d\ =\ \cancel \dfrac{7}{7}} \\ \\ \\ \sf : \implies {\purple{\underline{\boxed{\bf d\ =\ 1.}}}}\bigstar$

Hence:-

$\sf \therefore {\underline{The\ common\ difference\ of\ the\ given\ A.P\ is\ 1.}}$