If there are 9 AM's between 5 and 45 then sum of AM's will be

there are 9 ams between 5 and 45. i.e., by taking 5 as 1^st term 45 will be our 11^th term. so, 45=5+(11-1)d. then your d=4. now the sum of AMs will be 9Ams inserted 9[2*9+(9-1)4]/2=225. is it the answer? please reply me the correct answer.

 

the ams along with the nos. r in ap....so...t1=4 &  t11=45.45=5+10d.frm this d=4.t2=9 &t10=5+9*4=41

sum of all the ams are:9*(9+41)/2=9*25=225...this is the answer.