Question

please anybody solve this problem.​

Answer 1

Answer:

hey here is your answer

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Step-by-step explanation:

so let the required three digit number be

100(a-d)+10a+a+d ie nothing but

100a-100d+11a+d

so 111a-99d

wherein digit at hundreds place is a-d,at tens place is a and that of units is a+d

so as a-d,a,a+d are in A.P

the common difference between each of its terms would be constant

so here according to the given condition

a-d+a+a+d=21

ie 3a=21

so a=7

now according to second condition

100(a+d)+10a+a-d=100(a-d)+10a+a+d-396

ie 100(a+d)-d=100(a-d)+d-396

so 100a+100d=100a-100d+2d-396

ie 200d-2d=-396

ie 198d=-396

so d=-2

now just substitute these values in a-d,a,a+d

you get required digits as

a-d=9

a=7

a+d=5

so it assumed three digit number structure was

111a-99d

ie (111×7)-(9×(-2))

=777+198

=975

hence our required three digit number is 975

Answer 2

Step-by-step explanation:

So, the number is 111×5+99×3=852