It is given that an=13+6(n−1), which can be written as an=13+6n−6=7+6n Similarly, bn=15+7(n−1), which can be written as bn=15+7n−7=8+7n The common differences are 6 , and 7, respectively, The common difference of terms that exists in both series is I.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. tm=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+22×42=967