The Total solution for NCERT class 6-12
All the terms between10 and 250 are multiple of 4First multiple (a) = 12and last multiple (l) = 248and d = 4Let n be the number of multiples, thenan = a + (n – 1) d=> 248 = 12 + (n – 1) x 4 = 12 + 4n – 4=> 248 = 8 + 4n=> 4n = 248 – 8 = 240n = 60Number of terms are = 60