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

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, \dots, 407 and 2, 9, 16, \dots, 709

Commented by bobhans last updated on 19/Sep/20

l e t A_{n} = {3, 7, 11, \dots, 23, \dots, 51, \dots, 407}

B_{n} = {2, 9, 16, 23, \dots, 51, \dots, 709}

w e w a n t f i n d C_{n} = A_{n} \cap B_{n}

c o n s i d e r C_{n} = {23, 51, \dots}

C_{n} = 23 + (n - 1) . 28

C_{n} = 28 n - 5 < 407

28 n < 412; n = 14

t h e r e f o r e C_{n} = {23, 51, 79, \dots, 387}

Commented by bemath last updated on 19/Sep/20

s u p e r b \dots g a v e k u d o s

Commented by 1549442205PVT 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

{4 n}_{1} - 1 = {7 n}_{2} - 5

{7 n}_{2} - {4 n}_{1} = 4

{7 n}_{2} = 4 (n_{1} + 1)

∵ 4 ∣ n_{2} and 7 ∣ (n_{1} + 1)

\Rightarrow 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

\Rightarrow (n - 1) 28 ⩽ 384 \Rightarrow n ⩽ 14

∴ the no . of terms = 14

Commented by Rasheed.Sindhi last updated on 19/Sep/20

N i c e c i N

S i r i S

!

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

thank you sir .

Commented by 1549442205PVT 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, \dots

, 102 . Similarly, the terms of the second

A . P are obtained from the formula

b_{k} = 2 + 7 (k - 1), k = 1, 2, \dots, 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, 4 n + 4 = 7 k

From the equation 4 (n + 1) = 7 k it is

evident that k is a multiple of 4, that

is, k = 4 s with s = \overset{―}{1 \dots 25} (since 1 ⩽ k ⩽ 102)

. But if k = 4 s then 4 (n + 1) = 7 . 4 s, or

n = 7 s - 1 . Since 1 ⩽ n ⩽ 102, only the

numbers 1, 2, \dots, 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 = 7 s - 1, s = \overset{―}{1 \dots 14}, or from the

formula for b_{k}, for k = 4 s, s = \overset{―}{1 \dots 14}

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

great sir

great sir

Commented by 1549442205PVT last updated on 21/Sep/20

Thank Sir . You are welcome .

Thank Sir . You are welcome .