Answer :

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

T_{1} = 30 = a

T_{2} = 27

So, we have d = T_{2} – T_{1}

d = 27 – 30

d= – 3

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

And it is given that, S_{n} = 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 – 3n^{2}

⇒ 3n^{2} – 63n + 240 = 0

⇒ n^{2} – 21n + 80 = 0

⇒ n^{2} – 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 = T_{16}

So we have, T_{16} = a + (16 – 1)d ……. (∵ T_{n} = a + (n – 1)d )

⇒ T_{16} = a + 15d

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

⇒ T_{16}= 30 + 15(– 3)

⇒ T_{16} = 30 – 45

⇒ T_{16} = – 15

Now, when n=5, the last term = T_{5}

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

⇒ T_{5} = a + 4d

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

⇒ T_{5}= 30 + 4(– 3)

⇒ T_{5} = 30 – 12

⇒ T_{5} = 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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