In an arithmetic progression, if the sum of fourth, seventh and tenth terms is 99, and the sum of the first fourteen terms is 497, then the sum of first five terms is

The correct option is 65.

Let the first term of the arithmetic progression (AP) be a and its common difference be d.
The $n$-th term of an AP is given by the formula:
$a_n=a+(n-1)d$

The sum of the first $n$ terms of an AP is given by the formula:
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Step 1: Use the first given condition.
We are given that the sum of the fourth, seventh, and tenth terms is 99:
$a_4+a_7+a_{10}=99$
Expressing these terms in terms of a and d:
$(a+3d)+(a+6d)+(a+9d)=99$
Combining the like terms:
$3a+18d=99$
Dividing the entire equation by 3:
$a+6d=33$ — (Equation 1)

Step 2: Use the second given condition.
We are given that the sum of the first fourteen terms is 497:
$S_{14}=497$
Using the sum formula for $n=14$:
$\frac{14}{2}[2a+(14-1\left)d\right]=497$
$7[2a+13d]=497$
Dividing both sides by 7:
$2a+13d=71$ — (Equation 2)

Step 3: Solve the system of linear equations.
From Equation 1, we can express a as:
$a=33-6d$
Substitute this value of a into Equation 2:
$2(33-6d)+13d=71$
$66-12d+13d=71$
$66+d=71$
$d=71-66=5$
Now, substitute $d=5$ back into Equation 1 to find a:

Step 4: Find the sum of the first five terms.
We need to calculate $S_5$ with $a=3$ and $d=5$:
$S_5=\frac{5}{2}\left[2\right(3)+(5-1\left)\mathrm{(5)}\right]$
$S_5=\frac{5}{2}[6+4\times 5]$
$S_5=\frac{5}{2}[6+20]$
$S_5=\frac{5}{2}\times 26$
$S_5=5\times 13=65$

Thus, the sum of the first five terms is 65.