Let there be an A.P. with first term ‘a’, common difference, d. Ifdenotes its nth term andthe sum of first n terms, find.

(i)


(ii)


(iii)


(iv)


(v)


(vi)

(i)


Given, an= 50


a + (n – 1) d = 50


5 + (n – 1) 3 = 50


(n – 1)3 = 45


(n -1) = 15


n = 16


Sn= [a + l]


= 8 [5 + 50] = 8 [55] = 440


(ii)


an= 4, d = 2, Sn= -14


a + (n – 1) 2 = 4


a + 2n = 6, and


[2a + (n – 1)2] = -14


n [ 2a + 2n – 2] = -14, or


[a + an] = -14


[a + 4] = -14


n [6 -2n +4] = -28


n [10 -2n] = -28


2n2 – 10n – 28 = 0


2(n2 – 5n – 14) = 0


(n + 2) (n – 7) = 0


n = -2, n = 7


Therefore, n = -2 is not a natural number. So, n = 7


(iii)


Given, a = 3, n = 8, Sn= 192


Sn = {2a + (n – 1) d]


192 * 2 = 8 [6 + (8 – 1)d]


= 6 + 7d


48 = 6 + 7d


7d = 42


d = 6


(iv)


an= 28, Sn = 144, n = 9


Sn= [a + l]


144 = [a + 28]


= a = 28


a + 28 = 32


a = 4


(v)


Given, a = 8, an= 62 and Sn= 210


Sn= [a + l]


210 = [8 + 62]


210 * 2 = n (70)


n = = 6


a + (n – 1) d = 62


8 + (6 – 1) d = 62


5d = 54


d = 10.8


(vi)


Given, a = 2, d = 8 and Sn = 90


90 = [4 + (n – 1)d] {Therefore, Sn= [2a + (n – 1)d]}


180 = n [4 + 8n – 8]


8n2 – 4n – 180 = 0


4 (2n2 – n – 45) = 0


2n2 – n – 45 = 0


(2n + 1) (n – 5) = 0


Therefore, n = is not a natural number


n = 5


an= a + (n – 1)d


= 2 + 4(8) = 32


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