Question

If the sum of first $75$ terms of an A.P. is $2625$, then the $38th$ term of an A.P. is
1.$39$
2. $37$
3. $36$
4. $35$

Answer 1

Hint: In this question, we are given the sum of first $75$ terms of an A.P. is $2625$. First step is to apply the formula of Sum of $n$ terms in Arithmetic progression i.e., ${S_n} = \left( {\dfrac{n}{2}} \right)\left( {2a + \left( {n - 1} \right)d} \right)$. Put $n = 75$ here. In the end after solving completely you will get the $38th$ term of an A.P.. Also, the formula of the $nth$ term of A.P. is ${a_n} = a + \left( {n - 1} \right)d$.

Formula used:
The $nth$ term of A.P. is ${a_n} = a + \left( {n - 1} \right)d$
Sum of $n$ terms in Arithmetic progression –
${S_n} = \left( {\dfrac{n}{2}} \right)\left( {2a + \left( {n - 1} \right)d} \right)$
Where, $a$ is the first term of the series and $d$ is the common difference between each term.

Complete step by step solution: 
Using formula of $nth$ term of A.P. ${a_n} = a + \left( {n - 1} \right)d$
The $38th$ term of an A.P. is $a + 37d - - - - - \left( 1 \right)$
Given that,
The sum of first $75$ terms of an A.P. is $2625$
Using, formula of sum of $n$ terms in Arithmetic progression
${S_n} = \left( {\dfrac{n}{2}} \right)\left( {2a + \left( {n - 1} \right)d} \right) = 2625 - - - - - \left( 2 \right)$
Here, $n = 75$
Equation (2) will be,
$\left( {\dfrac{{75}}{2}} \right)\left( {2a + \left( {75 - 1} \right)d} \right) = 2625$
$\left( {2a + \left( {75 - 1} \right)d} \right) = 2625 \times \dfrac{2}{{75}}$
$2a + 74d = 70$
$2\left( {a + 37d} \right) = 70$
$a + 37d = 35$
From equation (1),
The $38th$ term of an A.P. is $35$.
Hence, Option (4) is the correct answer i.e., $35$.

Note: The key concept involved in solving this problem is the good knowledge of Arithmetic progression. Students must remember that arithmetic progression or arithmetic sequence is a number sequence in which the difference between consecutive terms is constant.