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 answer is 65.

Step 1: Understand the formulas for an Arithmetic Progression (A.P.)

Let the first term of the arithmetic progression be $a$ and the common difference be $d$.

The $n$-th term of an A.P. is given by the formula:

$a_n=a+(n-1)d$

The sum of the first $n$ terms of an A.P. is given by the formula:

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Step 2: Form the first equation using the given conditions

We are given that the sum of the 4th, 7th, and 10th terms is 99:

$a_4+a_7+a_{10}=99$

Expressing each term using the formula for the $n$-th term:

$a_4=a+3d$

$a_7=a+6d$

$a_{10}=a+9d$

Adding these three equations together:

$(a+3d)+(a+6d)+(a+9d)=99$

$3a+18d=99$

Dividing the entire equation by 3:

$a+6d=33$       -- (Equation 1)

Step 3: Form the second equation using the sum of the first fourteen terms

We are given that the sum of the first 14 terms ($S_{14}$) is 497:

$S_{14}=\frac{14}{2}[2a+(14-1\left)d\right]=497$

$7[2a+13d]=497$

Dividing both sides by 7:

$2a+13d=71$       -- (Equation 2)

Step 4: Solve Equation 1 and Equation 2 for $a$ and $d$

From Equation 1, express $a$ in terms of $d$:

$a=33-6d$

Substitute $a=33-6d$

$2(33-6d)+13d=71$

$66-12d+13d=71$

$66+d=71$

$d=71-66=5$

Now substitute $d=5$ back to find $a$:

$a=33-6\left(5\right)=33-30=3$

Step 5: Calculate the sum of the first five terms ($S_5$)

Using the sum formula for $n=5$:

$S_5=\frac{5}{2}[2a+(5-1\left)d\right]$

Substitute $a=3$ and $d=5$:

$S_5=\frac{5}{2}\left[2\right(3)+4(5\left)\right]$

$S_5=\frac{5}{2}[6+20]$

$S_5=\frac{5}{2}\left[26\right]$

$S_5=5\times 13=65$

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