Question

Find 4 numbers in an arithmetic progressions such that the sum of 1st and 3rd terms is 66 and the difference of 2nd and 4th terms is 22
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Answer 1

Answer:

Explanation:

Given that the difference of second and forth term is 22

So, if we divide by 2 then we get 11 so d is 11 ,

the sum of 1st and 3rd term is 66

We know the d is 11 and the sum of two terms are 66

If we substract 22 from 66 then we get 44 and divide by 2 we get 22

So first term is 22

22,33,44,55....