Question

the smallest positive integer that appears in each of the arithmetic progression 5,16,27,38,49,....; 7,20,33,46,59...; and 8,22,36,50,64.... is

Answer 1

The smallest positive integer appearing in all three progressions is 1996.

Step-by-step explanation:

The first AP is 5, 16, 27, 38, 49, ...

The mth term be

5 + 11 (n - 1) = 11m - 6

The second AP is 7, 20, 33, 46, 59, ...

The nth term be

7 + 13 (n - 1) = 13n - 6

The third AP is 8, 22, 36, 50, 64, ...

The pth term be

8 + 14 (n - 1) = 14p - 6

To find the smallest positive integer appearing in all three progressions, we equate the respective mth, nth and pth terms.

Then,

11m - 6 = 13n - 6 = 14p - 6

i.e., 11m = 13n = 14p

We have to find the smallest number divisible by 11, 13 and 14, we find the LCM of 11, 13 and 14, which is (11 * 13* 14 =) 2002.

Therefore the smallest positive integer appearing in all three progressions is (2002 - 6 =) 1996.

1. In an AP, if mth term is n and nth term is m, where m is not equal to n , find the pth term..

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2. If mth term of AP is 1/n and nth term is 1/m then find the sum of its first mn term.

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Answer 2

the smallest positive integer that appears in each of the AP is 1996

Step-by-step explanation:

number would be

11a + 5 = 13b + 7 = 14c + 8

11, 13 & 14 are co prime

and LCM =2002

so such numbers will repeat every 2002

one method to find using hit and trial

let's take 2 at a time

11a + 5 = 13b +7

11a = 13b + 2

b = 10 a = 12

b = 21 a = 25

and so on

13b + 7 = 14c + 8

13b = 14c + 8

c= 5. b = 6

c = 18. b = 20

11a+5 = 14c + 8

11a = 14c + 3

c = 10 a = 13

c = 21 a = 27

if we notice a sequence

13n -1. and 14n - 1

so common first a =181 then after 182

for this a = 181

b = 153. c = 142

now check

11a + 5 = 13b + 7 = 14c + 8

1996 = 1996 = 1996

1996 will be smallest integer satisfying this condition

another method

if we see series

5,16,27,38,49,....; 7,20,33,46,59...; and 8,22,36,50,64....

we see difference between numbers and difference between common difference is same

hence if we see one number back side. it will satish

5-11 = -6

7-13=-6

8-14=-6

LCMof 11,13& 14 = 2002

now add 2002 in -6 to get 1996