Let the first term, common difference of an AP are a and d, respectively.
According to the question,
a3+a8=7 and a7+a14=−3
⇒ a+(3−1)d+a+(8−1)d=7 [∵ an=a+(n−1)d]
And a + (7 - 1)d + a + (14 - 1)d = - 3
a + 2d + a + 7d = 7
And a + 6d + a + 13d =-3
2a + 9d = 7 . . . . . . . (i)
And 2a + 19d = - 3 . . . . . . . (ii)
On subtracting eq. (i) from eq.(ii), we get;
10d=−10⇒d=−1
2a + 9 (-1) = 7
⇒ 2a - 9 = 7
⇒ 2a = 16 ⇒ a = 8
∴ a10=a+(10−1)d
= 8 + 9(-1)
= 8 - 9 = -1