Question 22.

It is given that $a_n = 13 + 6(n-1)$, which can be written as $a_n = 13 + 6n - 6 = 7 + 6n$

Similarly, $b_n = 15 + 7(n-1)$, which can be written as $b_n = 15 + 7n - 7 = 8 + 7n$

The common differences are 6, and 7, respectively. The common difference of terms that exists in both series is l.c.m (6, 7) = 42

The first common term of the first two series is 43 (by inspection)

Hence, we need to find the mth term, which is less than 1000, and the largest three-digit integer, and exists in both series.

$t_m$ = a + (m-1)d < 1000

=> 43 + (m-1)42 < 1000

=> (m-1)42 < 957

=> m-1 < 22.8 => m < 23.8 => m = 23

Hence, the 23rd term is $43 + 22 \times 42 = 967$