












Courses
Q.
An A.P. 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 A.P.
see full answer
Start JEE / NEET / Foundation preparation at rupees 99/day !!
(Unlock A.I Detailed Solution for FREE)
Ready to Test Your Skills?
Check your Performance Today with our Free Mock Test used by Toppers!
Take Free Test
Detailed Solution
We are given the sum of the three middlemost terms of an A.P. and the sum of the last three terms, and we have to find the A.P.
The A.P. has 37 terms, so ๐ = 37.
Let the first term of the A.P. be ๐ and the common difference be ๐. The middlemost term of the A.P. is = ๐+1/2
37+1/ 2
38/ 2
= 19
So, the 19๐กโ term is the middlemost term.
The sum of the three middlemost terms = 225
The three middlemost terms are the 18๐กโ, 19๐กโ and 20๐กโ terms.
So, ๐18 + ๐19 + ๐20 = 225
โ ๐ + (18 โ 1)๐ + ๐ + (19 โ 1)๐ + ๐ + (20 โ 1)๐ = 225
โ 3๐ + 17๐ + 18๐ + 19๐ = 225
โ 3๐ + 54๐ = 225
โ 3(๐ + 18๐) = 225
โ ๐ + 18๐ = 225/3
โ ๐ + 18๐ = 75 ....(1)
The sum of the last three terms = 429
The three last terms are the 35๐กโ, 36๐กโ and 37๐กโ terms.
So, ๐35 + ๐36 + ๐37 = 429
โ ๐ + (35 โ 1)๐ + ๐ + (36 โ 1)๐ + ๐ + (37 โ 1)๐ = 429
โ 3๐ + 34๐ + 35๐ + 36๐ = 429
โ 3๐ + 105๐ = 429
โ 3(๐ + 35๐) = 429
โ ๐ + 35๐ = 429/3
โ ๐ + 35๐ = 143 ....(2)
Subtracting eq(1) from eq(2), we get
๐ + 35๐ โ (๐ + 18๐) = 143 โ 75
โ ๐ + 35๐ โ ๐ โ 18๐ = 68
โ 17๐ = 68
โ ๐ = 68/17
โ ๐ = 4
So, the common difference between the A.P. is 4. Putting the value of ๐ in eq(1), we get
๐ + 18 ร 4 = 75
โ ๐ + 72 = 75
โ ๐ = 75 โ 72
โ ๐ = 3
So, the first term of the A.P. is 3.
Hence, the A.P. can be written as ๐, ๐ + ๐, ๐ + 2๐, ๐ + 3๐,..., ๐ + 36๐
= 3, 7, 11, 15, ..., 147.
Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE