Question

An ap consists of 37 terms the sum of the three middle most terms is 225 and the sum of the other last key terms is 429

Answer 1

Here, is your answer.....

Answer 2

Answer:

3  , 7 , 11 , 15 & so on

Step-by-step explanation:

n  = 37

middle = (37+1)/2 = 19th terms

middle three terms  18th , 19th & 20th

18th term = a +17d

19th term = a +18d

20th term = a + 19d

Sum of middle 3 terms = 3a + 54d = 225

=> 3(a +18d) = 225

=> a +18d = 75    - Eq 1

Last three terms 37th , 36th & 35th

37th term = a +36d

36th Term = a +35d

35th term = a + 34d

Sum of last three terms = 3a+ 105d = 429

=> 3 (a + 35d) = 429

=> a + 35d = 143    - eq 2

Eq2 - Eq 1

17d = 68

d = 4

a + 18*(4) = 75

a = 3

AP is

3  , 7 , 11 , 15 & so on