For the A.P.: -3,-7,-11,..., can we find $a_{30} - a_{20}$ without actually finding $a_{30} a n d a_{20}$ ? Give reasons for your answer.

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Solution

$a_{n} - a_{k} = a + (n - 1) d - [a + (k - 1) d] a_{n} - a_{k} = (n - 1) d - (k - 1) d a_{n} - a_{k} = d [n - k] S o, a_{30} - a_{20} = - 4 (30 - 20) = - 4 \times 10 = - 40$

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