Question Number 114511 by 1549442205PVT last updated on 19/Sep/20

Find numbers which are common  terms of the two following arithmetic  progression:  3,7,11,...,407 and 2,9,16,...,709

Commented bybobhans last updated on 19/Sep/20

let A_n ={3,7,11,...,23,...,51,...,407}         B_n ={2,9,16,23,...,51,...,709}  we want find C_n = A_n ∩B_n   consider C_n ={23,51,... }                       C_n = 23+(n−1).28                        C_n =28n−5 < 407                                    28n < 412; n = 14  therefore C_n  = {23,51,79,...,387}

Commented bybemath last updated on 19/Sep/20

superb...gave kudos

Commented by1549442205PVT last updated on 19/Sep/20

Thank Sir.

Answered by PRITHWISH SEN 2 last updated on 19/Sep/20

3+(n_1 −1)4=2+(n_2 −1)7  4n_1 −1=7n_2 −5  7n_2 −4n_1 = 4  7n_2 = 4(n_1 +1)  ∵ 4∣n_2  and 7∣(n_1 +1)  ⇒n_1 =6 and n_2 = 4  ∴ the 1^(st) term is =4.6−1=7.4−5=23  Now we have to find an A.P whose 1^(st) term is 23  and c.d is = 4.7=28 and last term is ≤ 407  ∴t_n  = 23+(n−1)28≤407  ⇒ (n−1)28≤ 384⇒n≤14  ∴ the no. of terms = 14

Commented byRasheed.Sindhi last updated on 19/Sep/20

NiceciN    SiriS         !

Commented byPRITHWISH SEN 2 last updated on 19/Sep/20

thank you sir.

Commented by1549442205PVT last updated on 19/Sep/20

Thank sir,solution is pretty clear

Answered by 1549442205PVT last updated on 19/Sep/20

It is clear that the general term of the  first A.P is of the form a_n =3+4(n−1);  the indicated terms of the progression  are associated with the values n=1,2,...  ,102.Similarly,the terms of the second  A.P are obtained from the formula  b_k =2+7(k−1),k=1,2,...,102.The problem  thus consists in finding all numbers n  and k,1≤n≤102,1≤k≤102,for which   a_n =b_k ,that is,4n+4=7k  From the equation 4(n+1)=7k it is   evident that k is a multiple of 4,that  is ,k=4s with s=1...25^(−) (since 1≤k≤102)  .But if k=4s then 4(n+1)=7.4s,or  n=7s−1.Since 1≤n≤102,only the  numbers 1,2,...,14 are permissible   values of s.Thus,we have 14 numbers  that are common to both A.Ps  The numbers themselves are readily  found either from the formula for a_n   when n=7s−1,s=1...14^(−) ,or from the  formula for b_k ,for k=4s,s=1...14^(−)

Commented byPRITHWISH SEN 2 last updated on 19/Sep/20

great sir

Commented by1549442205PVT last updated on 21/Sep/20

Thank Sir.You are welcome.