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

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

We have:

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

$$\Rightarrow 3a + 18d = 99$$  or  $$a + 6d = 33$$ .....(1)

We are also told that the sum of the first fourteen terms is $$497$$,

$$\dfrac{14}{2}(2a + (14-1)d) = 497$$

$$\Rightarrow (2a+13d) = \dfrac{497}{7}$$

$$\Rightarrow 2a+13d = 71$$ .....(2)

Solving equations (1) and (2), we get;

$$a= 3$$ and $$d=5$$

The first five terms would therefore be: $$3, 8, 13, 18, 23$$, and their sum would be $$3+8+13+18+23 = 65$$.