The $17$ th term of an AP exceeds its $10$ th term by $7$ . Find the common difference

\frac{3}{7}

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\frac{- 3}{7}

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\frac{7}{3}

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\frac{- 7}{3}

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Solution

The correct option is C $\frac{7}{3}$
$a_{7} = a + 6 d$
And, $a_{10} = a + 9 d$
Or, $a + 9 d - a - 6 d = 7$
Or, $3 d = 7$
Or, $d = 7 / 3$

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