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

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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
$$\mathrm{Find}\:\mathrm{numbers}\:\mathrm{which}\:\mathrm{are}\:\mathrm{common} \\ $$$$\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{following}\:\mathrm{arithmetic} \\ $$$$\mathrm{progression}: \\ $$$$\mathrm{3},\mathrm{7},\mathrm{11},…,\mathrm{407}\:\mathrm{and}\:\mathrm{2},\mathrm{9},\mathrm{16},…,\mathrm{709} \\ $$
Commented by bobhans 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}
$${let}\:{A}_{{n}} =\left\{\mathrm{3},\mathrm{7},\mathrm{11},…,\mathrm{23},…,\mathrm{51},…,\mathrm{407}\right\} \\ $$$$\:\:\:\:\:\:\:{B}_{{n}} =\left\{\mathrm{2},\mathrm{9},\mathrm{16},\mathrm{23},…,\mathrm{51},…,\mathrm{709}\right\} \\ $$$${we}\:{want}\:{find}\:{C}_{{n}} =\:{A}_{{n}} \cap{B}_{{n}} \\ $$$${consider}\:{C}_{{n}} =\left\{\mathrm{23},\mathrm{51},…\:\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{C}_{{n}} =\:\mathrm{23}+\left({n}−\mathrm{1}\right).\mathrm{28} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{C}_{{n}} =\mathrm{28}{n}−\mathrm{5}\:<\:\mathrm{407}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{28}{n}\:<\:\mathrm{412};\:{n}\:=\:\mathrm{14} \\ $$$${therefore}\:{C}_{{n}} \:=\:\left\{\mathrm{23},\mathrm{51},\mathrm{79},…,\mathrm{387}\right\} \\ $$
Commented by bemath last updated on 19/Sep/20
superb...gave kudos
$${superb}…{gave}\:{kudos} \\ $$
Commented by 1549442205PVT last updated on 19/Sep/20
Thank Sir.
$$\mathrm{Thank}\:\mathrm{Sir}. \\ $$
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
$$\mathrm{3}+\left(\mathrm{n}_{\mathrm{1}} −\mathrm{1}\right)\mathrm{4}=\mathrm{2}+\left(\mathrm{n}_{\mathrm{2}} −\mathrm{1}\right)\mathrm{7} \\ $$$$\mathrm{4n}_{\mathrm{1}} −\mathrm{1}=\mathrm{7n}_{\mathrm{2}} −\mathrm{5} \\ $$$$\mathrm{7n}_{\mathrm{2}} −\mathrm{4n}_{\mathrm{1}} =\:\mathrm{4} \\ $$$$\mathrm{7n}_{\mathrm{2}} =\:\mathrm{4}\left(\mathrm{n}_{\mathrm{1}} +\mathrm{1}\right) \\ $$$$\because\:\mathrm{4}\mid\mathrm{n}_{\mathrm{2}} \:\boldsymbol{\mathrm{and}}\:\mathrm{7}\mid\left(\boldsymbol{\mathrm{n}}_{\mathrm{1}} +\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{n}_{\mathrm{1}} =\mathrm{6}\:\mathrm{and}\:\mathrm{n}_{\mathrm{2}} =\:\mathrm{4} \\ $$$$\therefore\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \mathrm{term}\:\mathrm{is}\:=\mathrm{4}.\mathrm{6}−\mathrm{1}=\mathrm{7}.\mathrm{4}−\mathrm{5}=\mathrm{23} \\ $$$$\mathrm{Now}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{find}\:\mathrm{an}\:\mathrm{A}.\mathrm{P}\:\mathrm{whose}\:\mathrm{1}^{\mathrm{st}} \mathrm{term}\:\mathrm{is}\:\mathrm{23} \\ $$$$\mathrm{and}\:\mathrm{c}.\mathrm{d}\:\mathrm{is}\:=\:\mathrm{4}.\mathrm{7}=\mathrm{28}\:\mathrm{and}\:\mathrm{last}\:\mathrm{term}\:\mathrm{is}\:\leqslant\:\mathrm{407} \\ $$$$\therefore\boldsymbol{\mathrm{t}}_{\boldsymbol{\mathrm{n}}} \:=\:\mathrm{23}+\left(\boldsymbol{\mathrm{n}}−\mathrm{1}\right)\mathrm{28}\leqslant\mathrm{407} \\ $$$$\Rightarrow\:\left(\boldsymbol{\mathrm{n}}−\mathrm{1}\right)\mathrm{28}\leqslant\:\mathrm{384}\Rightarrow\mathrm{n}\leqslant\mathrm{14} \\ $$$$\therefore\:\mathrm{the}\:\mathrm{no}.\:\mathrm{of}\:\mathrm{terms}\:=\:\mathrm{14} \\ $$$$ \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 19/Sep/20
NiceciN SiriS !
$$\mathcal{N}{iceci}\mathcal{N} \\ $$$$\:\:\mathcal{S}{iri}\mathcal{S} \\ $$$$\:\:\:\:\:\:\:! \\ $$
Commented by PRITHWISH SEN 2 last updated on 19/Sep/20
thank you sir.
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by 1549442205PVT last updated on 19/Sep/20
Thank sir,solution is pretty clear
$$\mathrm{Thank}\:\mathrm{sir},\mathrm{solution}\:\mathrm{is}\:\mathrm{pretty}\:\mathrm{clear} \\ $$
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^(−)
$$\mathrm{It}\:\mathrm{is}\:\mathrm{clear}\:\mathrm{that}\:\mathrm{the}\:\mathrm{general}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{first}\:\mathrm{A}.\mathrm{P}\:\mathrm{is}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}\:\mathrm{a}_{\mathrm{n}} =\mathrm{3}+\mathrm{4}\left(\mathrm{n}−\mathrm{1}\right); \\ $$$$\mathrm{the}\:\mathrm{indicated}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{progression} \\ $$$$\mathrm{are}\:\mathrm{associated}\:\mathrm{with}\:\mathrm{the}\:\mathrm{values}\:\mathrm{n}=\mathrm{1},\mathrm{2},… \\ $$$$,\mathrm{102}.\mathrm{Similarly},\mathrm{the}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{second} \\ $$$$\mathrm{A}.\mathrm{P}\:\mathrm{are}\:\mathrm{obtained}\:\mathrm{from}\:\mathrm{the}\:\mathrm{formula} \\ $$$$\mathrm{b}_{\mathrm{k}} =\mathrm{2}+\mathrm{7}\left(\mathrm{k}−\mathrm{1}\right),\mathrm{k}=\mathrm{1},\mathrm{2},…,\mathrm{102}.\mathrm{The}\:\mathrm{problem} \\ $$$$\mathrm{thus}\:\mathrm{consists}\:\mathrm{in}\:\mathrm{finding}\:\mathrm{all}\:\mathrm{numbers}\:\mathrm{n} \\ $$$$\mathrm{and}\:\mathrm{k},\mathrm{1}\leqslant\mathrm{n}\leqslant\mathrm{102},\mathrm{1}\leqslant\mathrm{k}\leqslant\mathrm{102},\mathrm{for}\:\mathrm{which}\: \\ $$$$\mathrm{a}_{\mathrm{n}} =\mathrm{b}_{\mathrm{k}} ,\mathrm{that}\:\mathrm{is},\mathrm{4n}+\mathrm{4}=\mathrm{7k} \\ $$$$\mathrm{From}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{4}\left(\mathrm{n}+\mathrm{1}\right)=\mathrm{7k}\:\mathrm{it}\:\mathrm{is}\: \\ $$$$\mathrm{evident}\:\mathrm{that}\:\mathrm{k}\:\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{4},\mathrm{that} \\ $$$$\mathrm{is}\:,\mathrm{k}=\mathrm{4s}\:\mathrm{with}\:\mathrm{s}=\overline {\mathrm{1}…\mathrm{25}}\left(\mathrm{since}\:\mathrm{1}\leqslant\mathrm{k}\leqslant\mathrm{102}\right) \\ $$$$.\mathrm{But}\:\mathrm{if}\:\mathrm{k}=\mathrm{4s}\:\mathrm{then}\:\mathrm{4}\left(\mathrm{n}+\mathrm{1}\right)=\mathrm{7}.\mathrm{4s},\mathrm{or} \\ $$$$\mathrm{n}=\mathrm{7s}−\mathrm{1}.\mathrm{Since}\:\mathrm{1}\leqslant\mathrm{n}\leqslant\mathrm{102},\mathrm{only}\:\mathrm{the} \\ $$$$\mathrm{numbers}\:\mathrm{1},\mathrm{2},…,\mathrm{14}\:\mathrm{are}\:\mathrm{permissible}\: \\ $$$$\mathrm{values}\:\mathrm{of}\:\mathrm{s}.\mathrm{Thus},\mathrm{we}\:\mathrm{have}\:\mathrm{14}\:\mathrm{numbers} \\ $$$$\mathrm{that}\:\mathrm{are}\:\mathrm{common}\:\mathrm{to}\:\mathrm{both}\:\mathrm{A}.\mathrm{Ps} \\ $$$$\mathrm{The}\:\mathrm{numbers}\:\mathrm{themselves}\:\mathrm{are}\:\mathrm{readily} \\ $$$$\mathrm{found}\:\mathrm{either}\:\mathrm{from}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{a}_{\mathrm{n}} \\ $$$$\mathrm{when}\:\mathrm{n}=\mathrm{7s}−\mathrm{1},\mathrm{s}=\overline {\mathrm{1}…\mathrm{14}},\mathrm{or}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{formula}\:\mathrm{for}\:\mathrm{b}_{\mathrm{k}} ,\mathrm{for}\:\mathrm{k}=\mathrm{4s},\mathrm{s}=\overline {\mathrm{1}…\mathrm{14}} \\ $$
Commented by PRITHWISH SEN 2 last updated on 19/Sep/20
great sir
$$\mathrm{great}\:\mathrm{sir} \\ $$
Commented by 1549442205PVT last updated on 21/Sep/20
Thank Sir.You are welcome.
$$\mathrm{Thank}\:\mathrm{Sir}.\mathrm{You}\:\mathrm{are}\:\mathrm{welcome}. \\ $$

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