Question

Question 4
An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the AP.

Answer 1

Total number of terms (n) = 37
$\therefore Middleterm=\left(\frac{37+1}{2}\right)thterm={19}^{th}term$
So, the three middle most terms are ${18}^{th}$, ${19}^{th}$ and ${20}^{th}$ terms.
By the given condition,
Sum of the three middle most terms = 225
$a_{18}+a_{19}+a_{20}=225$
$\Rightarrow (a+17d)+(a+18d)+(a+19d)=225$
$\Rightarrow 3a+54d=225$
$\Rightarrow a+18d=75$ ...(i)
And,
Sum of the last three terms = 429
$\Rightarrow a_{35}+a_{36}+a_{37}=429$
$\Rightarrow (a+34d)+(a+35d)+(a+36d)=429$
$\Rightarrow 3a+105d=429$
$\Rightarrow a+35d=143$ ...(ii)
On subtracting Eq. (i) from Eq.(ii), we get,
17d = 68
$\Rightarrow$ d = 4
From Eq.(i),
a + 18(4) = 75
$\Rightarrow a=75-72$
$\Rightarrow a=3$
$\therefore$ Required AP is a, a + d, a + 2d, a + 3d
i.e., 3, 3+4, 3+2(4), 3+3(4)
i.e., 3, 7, 3+8, 3+12….
i.e., 3, 7, 11, 15….