In an A.P., the sum of first terms is – 150 and the sum of its next tem is-550. Find the A.P.

Sum of 10th term, S10 = -150

Sum of next 10 term = -550


Let, first term = a


Last term = an


We know, an = a + (n – 1) d


= a + (10 -1) d


= a + 9d


We know, Sum of n terms


Sn = [a + l]


S10 = [a + a + 9d]


-150 = 5 (2a + 9d)


-150 = 5 (2a + 9d)


10a = -150 – 45d


a = (i)


For next 10 term, First term = a11


Last term = a20


11th term, a11 = a + 10d


20th term, a20 = a +19d


S’10 = [a + 10d + a + 19d]


-550 = 5 [2a + 29d]


-550 = 10a + 145d


10a = -550 – 145d


a = (ii)


From (i) and (ii), we get


=


-150 – 45d = -550 – 145d


-45d = 145d = -550 + 150


100d = -400


d = -4


Put the value of d in (i), we get


a =


= = = 3


a1 = 3


a2 = a + d = 3 + (-4) = -1


a3 = a + 2d = 3 + 2(-4) = -5


Hence, A.P. is 3, -2, -5,…


And a = 3, d = -4


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