Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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.

$$\mathrm{Thank}\:\mathrm{Sir}.\mathrm{You}\:\mathrm{are}\:\mathrm{welcome}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com