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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 $3,\ 7,\ 11,\ 15\ …$

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