Question 1177527
Let first term = a and common difference = d
S_10=10/2 [2a+(10-1)d]=80
                                                                        (2a+9d)=16            ---------------- (i)

Also sum of next 12 terms is 624
Hence sum of first 22 terms = 624+80 = 704
Therefore, S_22=  22/2 [2a+(22-1)d]=704
                            2a + 21d = 64                ------------------------(ii)
 On solving (i) and (ii)
(2a + 21d)  -  (2a+9d) = 704  -  16
12d = 48
d = 4
from equation (i)  2a + 36 = 16
a= -10

Hence required AP is
-10, -6, -2, 2, 6, 10, 12, …………..