The sum of the second and the fifth term of an arithmetic progression (A.P) is $8$ and that of the third and the seventh term is $14$. Find the progression.
Let the first term of A.P $= t_1 = a$
Let the common difference of A.P $= d$
$n^{th}$ term of A.P $= t_n = a + \left(n - 1\right) d$
$\therefore \;$ $2^{nd}$ term of A.P $= t_2 = a + d$
$3^{rd}$ term of A.P $= t_3 = a + 2d$
$5^{th}$ term of A.P $= t_5 = a + 4d$
$7^{th}$ term of A.P $= t_7 = a + 6d$
Given: $\;$ $t_2 + t_5 = 8$
i.e. $\;$ $a + d + a + 4d = 8$
i.e. $\;$ $2a + 5d = 8$ $\;\;\; \cdots \; (1)$
And $\;$ $t_3 + t_7 = 14$
i.e. $\;$ $a + 2d = a + 6d = 14$
i.e. $\;$ $2a + 8d = 14$ $\;\;\; \cdots \; (2)$
Subtracting equations $(1)$ and $(2)$ gives
$3d = 6$ $\implies$ $d = 2$
Substituting $d = 2$ in equation $(1)$ gives
$2a + 5 \times 2 = 8$
i.e. $\;$ $2a = -2$ $\implies$ $a = -1$
$\therefore \;$ The required A.P is $\;$ $a, \; a+d, \; a+2d, \; a+3d, \; a+4d, \; a+5d, \; a+6d, \; \cdots$
i.e. $\;$ $-1, \; 1, \; 3, \; 5, \; 7, \; 9, \; 11, \; \cdots$