Find the sum :

(i)


(ii)


(iii)


(iv)


(v)


(vi)


(vii)


(viii)

(i)


a = 2, d = 4 – 2 = 2, l = 200


n = + 1


= + 1 = 100


S100= [2(2) + 99(2)]


= 100 [101] = 10100


(ii)


a = 3, d = 11 – 3 = 8, l = 803


Sn= + 1 = + 1


= 101


S101= [2(3) + 100(8)]


= 101 (403) = 40703


(iii)


a = -5, d = -8 + 5 = -3, l = -230


n = + 1


= = 76


Sn= 38(5) [-2 – 45]


= 190 (-47) = -8930


(iv)


a = 1, d = 3 – 1 = 2, l = 199


n = + 1


= 100


S100= [2(1) + 99(2)]


= 100 (100) = 10000


(v)


a = 7, d = - 7 =


Last term, an = 84


an = a + (n – 1)d


84 = 7 + (n – 1)


84 =


84 * 2 = 7 + 7n


168 = 7 + 7n


7n = 168 – 7


7n = 161


n = 23


Sum of nth term, Sn = [2a + (n – 1)d]


S23 = [2(7) + (23 – 1)]


= [14 + 22 * ]


= [14 + 77]


= * 91 =


Hence, sum of given A.P. is


(vi)


a = 34, d = 32 – 34 = -2


Last term, an = 10


an = a + (n – 1)d


10 = 34 + (n – 1) (-2)


10 = 34 – 2n + 2


2n = 34 – 10 + 2


2n = 26


n = 13


Sum of n terms,


Sn = [2a + (n – 1) d]


S13 = [2(34) + (13 – 1) (-2)]


= [68 + 12(-2)]


= [68 – 24]


= * 44 = 286


Hence, Sum of given A.P. is 286


(vii)


a = 25, d = 3


Last term, an = 100


An = a + (n -1) d


100 = 25 (n – 1)3


75 = 3n – 3


78 = 3n


n = 26


Sum of n terms, Sn = [2a + (n – 1) d]


= [2(25) + (26 – 1) 3]


= 13 [50 + 25 * 3]


= 13 * 125 = 1625


Hence, Sum of given A.P. is 1625


(viii)


a = 18, d = - 18 =


Last term, an =


an = a+ (n – 1)d


= 18 + (n – 1) ()


-99 = 36 + 5 – 5n


-140 = -5n


n = 28


S28= [a + l]


= 14 (18 - )


= 7 (36 – 99)


= 7 (-63) = -441


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