Answer :

We can see that, A.P. is 30, 27, 24, 21 …


T1 = 30 = a


T2 = 27


So, we have d = T2 – T1


d = 27 – 30


d= – 3


Now, we have sum of the A.P. Sn = × (2a + (n – 1)d)


And it is given that, Sn = 120


So, we have,


120 = × (2a + (n – 1)d)


Now we will put the values of a and d in the above equation


120 = × (2(30) + (n – 1)(– 3))


120 = × (60 + 3 – 3n)


120 × 2 = n × (63 – 3n)


240 = 63n – 3n2


3n2 – 63n + 240 = 0


n2 – 21n + 80 = 0


n2 – 16n – 5n + 80 = 0


n(n – 16) – 5(n – 16) = 0


(n – 16)(n – 5) = 0


n= 16 or 5


Now, when n=16, the last term = T16


So we have, T16 = a + (16 – 1)d ……. ( Tn = a + (n – 1)d )


T16 = a + 15d


Now we put the values of a and d in the above equation


T16= 30 + 15(– 3)


T16 = 30 – 45


T16 = – 15


Now, when n=5, the last term = T5


So we have, T5 = a + (5 – 1)d ……. ( Tn = a + (n – 1)d )


T5 = a + 4d


Now we put the values of a and d in the above equation


T5= 30 + 4(– 3)


T5 = 30 – 12


T5 = 18


The no. Of term in the above A.P. can be n = 5 or 16


And the last terms of the respective A.Ps are:


For n = 5, last term T5 = 18


For n = 16, last term T16 = – 15


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