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# id a,b,c are natural numbers and are in arithmetic progressions and a+b+c=21.then find the possible values for a,b,c

58 Points
8 years ago

sincea,b,c are in ap then b=a+c/2

so putting this we get

a+c=2b

so 2b+b=21 ,this means 3b=21

b=7

simple solve for a and c and u will get 6 and 8.

all the best!!!!!!!

souvik sonu roy
34 Points
8 years ago

a+c=2b

3b=21

b=7

a+c=14

a can be 1,2,3,4,5,6

c can be13,12,11,10,9,8

souvik sonu roy
34 Points
8 years ago

a+b+c=21

a+c=2b

3b=21

b=7

a can be 1,2,3,4,5,6

c can be 13,12,11,10,9,8

Manas Satish Bedmutha
22 Points
8 years ago
let a=b-d and c=b+d. Thus b=7. To find a+c=14, solutions, consider a=1 to 14 and crrespondingly c= 13 to 0. but 0 is not natural. So consider a= 1 to 13. Thus ther are 13 solutions, viz. (1,7,13)(2,7,12)(3,7,11)(4,7,10)(5,7,9)(6,7,8)(7,7,7)(8,7,6)(9,7,5)(10,7,4)(11,7,3)(12,7,2)(13,7,1)