Question

An AP consists of 37 terms .The sum of the three middlemost terms is 225 & the sum of the last three terms is 429 .Find the AP.

Answer 1

Given that an AP consists of 37 terms = > n = 37.

Then the middle-most term = (n + 1)/2

                                               = (37 + 1)/2

                                               = 38/2

                                               = 19.



The three middle most terms = a18,a19,a20.

Then the sum of the three middle most terms = a18 + a19 + a20.

                                                                              = (a + 17d) + (a+18d) + (a+19d).



Given that the sum of the three middlemost terms = 225.

a + 17d + a + 18d + a + 19d = 225

3a + 54d = 225

a + 18d = 75   ------------ (1).



Given that the total number of terms = 37.

The last 3 terms = a35,a36,a37.

Then the sum of last three terms = a35 +a36 + a37.

                                                        = (a + 34d) + (a+35d) + (a+36d)


Given that sum of last three terms = 429.

a + 34d + a + 35d + a+36d = 429

3a + 105d = 429

a + 35d = 143     ---------------- (2)


On solving (1) & (2), we get

a + 35d = 143

a + 18d = 75
--------------------

      17d = 68

          d = 4.

Substitute d = 4 in (1), we get

a + 18d = 75

a + 18(4) = 75

a = 75 - 72

a = 3.


Therefore the AP is 3,7,11,15,19,23......



Hope this helps!

Answer 2

plz refer to this attachment