SOLUTION: The sum of the first 10 terms of an arithmetic progression is 80 and the sum of the next 12 terms is 624. What is the arithmetic progression? Thank you

Question 1177527: The sum of the first 10 terms of an arithmetic progression is 80 and the sum of the next 12 terms is 624. What is the arithmetic progression?
Thank you
Found 2 solutions by jitendra_maths, mananth:
Answer by jitendra_maths(6) (Show Source): You can put this solution on YOUR website!
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, …………..

Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
The sum of the first 10 terms of an arithmetic progression is 80 and the sum of the next 12 terms is 624. What is the arithmetic progression?
S10 = 10/2(2a+(10-1)d)
S10 = 10a+45d =80 ----------------1
S22 = 22/2(2a+(22-1)d)
S22 = 11(2a+21d)
S22= 22a+231d
S22-S10 =80
22a+231d-80 =624
22a +231d = 704
/11
2a+21d=64----------------2
solve equation 1 & 2
a=-10,d=4
-10,-6,-2,2,6,...............

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