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

S_n= + 1 = + 1

= 101

S₁₀₁= [2(3) + 100(8)]

= 101 (403) = 40703

(iii)

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

n = + 1

= = 76

S_n= 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, a_n = 84

a_n = 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 n^th term, S_n = [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, a_n = 10

a_n = 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,

S_n = [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, a_n = 100

A_n = a + (n -1) d

100 = 25 (n – 1)3

75 = 3n – 3

78 = 3n

n = 26

Sum of n terms, S_n = [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, a_n =

a_n = 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