Find the first term $a_1$ and the common difference $d$ of the arithmetic progression (A.P) in which
$a_2 + a_5 - a_3 = 10$, $\;\;$ $a_2 + a_9 = 17$
First term of A.P $= a_1$, $\;$ Common difference of A.P $= d$
$n^{th}$ term of A.P $= a_n = a + \left(n - 1\right) d$
$\therefore \;$ $2^{nd}$ term of A.P $= a_2 = a_1 + d$
$3^{rd}$ term of A.P $= a_3 = a_1 + 2d$
$5^{th}$ term of A.P $= a_5 = a_1 + 4d$
$9^{th}$ term of A.P $= a_9 = a_1 + 8d$
Given: $\;$ $a_2 + a_5 - a_3 = 10$
i.e. $\;$ $a_1 + d + a_1 + 4d - a_1 - 2d = 10$
i.e. $\;$ $a_1 + 3d = 10$ $\;\;\; \cdots \; (1)$
and $\;$ $a_2 + a_9 = 17$
i.e. $\;$ $a_1 + d + a_1 + 8d = 17$
i.e. $\;$ $2a_1 + 9d = 17$ $\;\;\; \cdots \; (2)$
Solving equations $(1)$ and $(2)$ simultaneously, we have
Equation $(1) \times 2 \implies$ $2a_1 + 6d = 20$ $\;\;\; \cdots \; (3)$
Equation $(2) \times 1 \implies$ $2a_1 + 9d = 17$ $\;\;\; \cdots \; (4)$
Subtracting equations $3$ and $4$ gives
$-3 d = 3$ $\implies$ $d = -1$
Substitute $d = -1$ in equation $(1)$ to get
$a_1 - 3 = 10$ $\implies$ $a_1 = 13$