Question 24
An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three terms is 429. Find the AP.
Given
Number of terms of AP = 37
So, last term will be 37th term
Middle most term = ((37 + 1))/2 = 38/2 = 19th term
Theory
Let AP be
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
Here number of terms = 11
Middle term = 6
So, Middle term = ((11 + 1))/2 = 12/2 = 6
So, 19th term is middle most term
Thus,
3 middle most terms will be 18th term, 19th term, 20th term
18th term = a18
= a + (18 – 1)d
= a + 17d
19th term = a19
= a + (19 – 1)d
= a + 18d
20th term = a20
= a + (20 – 1)d
= a + 19d
Given
Sum of 3 middle most terms = 225
(a + 17d) + (a + 18d) + (a + 19d) = 225
(a + a + a) + (17d + 18d + 19d) = 225
3a + 54d = 225
3(a + 18d) = 225
a + 18d = 225/3
a + 18d = 75
Also,
Sum of last 3 terms = 429
Last term = an
= a37
= a + (37 – 1)d
= a + 36d
Second Last term = an – 1
= a37 – 1
= a36
= a + (36 – 1)d
= a + 35d
Third Last term = an – 2
= a37 – 2
= a35
= a + (35 – 1)d
= a + 34d
Now
Sum of last 3 terms = 429
Putting values
(a + 36d) + (a + 35d) + (a + 34d) = 429
(a + a + a) + (36d + 35d + 34d) = 429
3a + 105d = 429
3(a + 35d) = 429
a + 35d = 429/3
a + 35d = 143
Thus, our equations are
a + 18d = 7 5 …(1)
a + 35d = 143 …(2)
Doing (2) – (1)
(a + 35d) – (a + 18d) = 143 – 75
a – a + 35d – 18d = 143 – 75
17d = 68
d = 68/17
d = 4
From (1)
a + 18d = 7 5
Putting d = 4
a + 18 × 4 = 75
a + 72 = 75
a = 75 – 72
a = 3
Thus, a = 3, d = 4
So, AP is
3, 7, 11, 15, 19, …… 147

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.