Question

4. Find the Sn for the following arithmetic series described.
(i) a = 5, n = 30, l = 121 (ii) a = 50, n = 25, d =- 4

Answer 1

The sum of n terms of an AP is

S(n) =(n/2) (a+l) or

S(n) = (n/2) [2a + (n-1) d]

1. a=5, n=30, l =121

S(n) =(n/2) (a+l)

= (30/2) (5+121)

= 15 × 126

= 1890

2.
a= 50, n= 25, d= -4

S(n) = (n/2) [2a + (n-1) d]

= (25/2) [ 2(50) +(25-1)(-4)]

= (25/2) [ 100 +(24)(-4)]

= (25/2) [ 100 - 96]

= (25/2) [ 4]

= 25 × [2]

= 50

Answer 2

Sum of n terms of an AP
The sum of first n terms of an AP with first term 'a' and common difference 'd' is given by

Sn = n /2 [ 2a + ( n - 1) d] or
Sn=n /2 [ a + l ] (l = last term)

SOLUTION :
i) GIVEN : a= 5 , n = 30, l = 121

Sn= (n/2) [ a+ l ]
S30= (30/2) [ 5 + 121 ]
S30 = (15) [126]
S30 = 1890

(ii) Given : a = 50, n = 25 , d = - 4

Sn = (n/2) [ 2a + (n-1) d ]
S₂₅ = (25/2) [ 2(50) + (25-1) (- 4) ]
S₂₅ = (25/2) [ 100 + (24) (-4) ]
S₂₅ = (25/2) [ 100 - 96 ]
S₂₅ = (25/2) [4]
S₂₅ = (25) [2]
S₂₅ = 50

HOPE THIS WILL HELP YOU...