difference of AP is 4.(b) First term of AP is 5.Explanation:Given information,Fifth term of an arithmetic sequence is 21 and it's ninth term is 37.(a) what is it's COMMON difference?(b) What is it's first term?Here,\SF a_{5}a 5 = 21\sf a_{9}a 9 = 37a = ?d = ?⚘ Using formula of nth term ::\bf{\dag}\:{\boxed{\tt{a_{n} = a + (n - 1)d}}}† a n =a+(n−1)d ⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━(a)➨ \tt a_{5} = a + (5 - 1)da 5 =a+(5−1)d➨ \tt 21 = a + 4d21=a+4d➨ \tt a = 21 - 4d\qquad- (1)a=21−4d−(1)Also,➨ \tt a_{9} = a + (9 - 1)da 9 =a+(9−1)d➨ \tt 37 = a + 8d37=a+8d➨ \tt a = 37 - 8d\qquad- (2)a=37−8d−(2)From (1) & (2) we get,➨ \tt 21 - 4d = 37 - 8d21−4d=37−8d➨ \tt 21 - 37 = - 8d + 4d21−37=−8d+4d➨ \tt \cancel{-} 16 = \cancel{-} 4d − 16= − 4d➨ \tt 4d = 164d=16➨ \tt d = {\cancel{\dfrac{16}{4}}}d= 416 ➨ d = 4Hence, common difference (d) of AP is 4.(b)PUT d = 4 in (1) we get,➨ \tt a = 21 - (4\:\times\:4)a=21−(4×4)➨ \tt a = 21 - 16a=21−16➨ a = 5Hence, first term (a) of AP is 5.Verification:➨ \tt a_{5} = a + (5 - 1)da 5 =a+(5−1)d➨ \tt 21 = a + 4d21=a+4dBy putting value of a and d in above equation we get,➨ \tt 21 = 5 + (4\:\times\:4)21=5+(4×4)➨ \tt 21 = 5 + 1621=5+16➨ \tt 21 = 2121=21➨ LHS = RHSAlso,➨ \tt a_{9} = a + (9 - 1)da 9 =a+(9−1)d➨ \tt 37 = a + 8d37=a+8dBy putting value of a and d in above equation we get,➨ \tt 37 = 5 + (8\:\times\:4)37=5+(8×4)➨ \tt 37 = 5 + 3237=5+32➨ \tt 37 = 3737=37➨ LHS = RHSHence, Verified ✔▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬