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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.
Given:
An A.P. consists of $37$ terms. The sum of the three middle most terms is $225$ and the sum of the last three is $429$.
To do:
We have to find the A.P.
Solution:
The A.P contains $37$ terms. So, the middle most term is $( \frac{37+1}{2})^{th}$ term $= 19^{th}$ term.
Thus, three middle most terms of this A.P. are $18^{th}$, $19^{th}$ and $20^{th}$ terms.
Given $a_{18} + a_{19} + a_{20} = 225$
$\Rightarrow (a + 17d) + (a + 18d) + (a + 19d) = 225$
$\Rightarrow 3(a + 18d) = 225$
$\Rightarrow a + 18d = 75$
$\Rightarrow a = 75 – 18d\ … ( 1)$
As given:
$a_{35} + a_{36} + a_{37} = 429$
$\Rightarrow ( a + 34d) + ( a + 35d) + ( a + 36d) = 429$
$\Rightarrow 3( a + 35d) = 429$
$\Rightarrow ( 75 – 18d) + 35d = 143$
$\Rightarrow 17d = 143 – 75 = 68$
$\Rightarrow d = 4$
Substituting the value of $d$ in equation $( 1)$, we get,
$a = 75 – 18 \times 4 = 3$
Thus, the required A.P. is $3,\ 7,\ 11,\ 15\ …$