If the 17ᵗʰ term of an arithmetic progression is 7 more than its 10ᵗʰ term, then the common difference of that arithmetic progression is.

Explanation

From the first condition, $T_{17} = a + (17 - 1) d$ $T_{17} = a + 16d.......................$ (i) From the second condition, $T_{10} = a + (10 - 1) d$ $T_{10} = a + 9d.......................$ (ii) As per the question, $a + 16 d - a - 9d = 7$ $7d = 7$ d = 1 Therefore, the common difference of the series is 1.