Question

6. In an arithmetic series, the sum of first 11 terms is 44 and that of the next 11 terms is
55. Find the arithmetic series.

Answer 1

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)

General form of an AP.:
a, a+d, a+2d, a+3d…….

SOLUTION :
Given :
Sum of first 11 terms = 44
Sum of next 11 term = 55
S₁₁ = 44
Sn = n /2 [ 2a + ( n - 1) d]
44 = (11/2)[2a + (11-1) d]
44 = (11/2)[2a + 10d]
2a + 10 d = (44 x 2)/11
2a + 10 d = 4 x 2
2a + 10 d = 8 . …. …. (1)

Sum of next 11 term = 55
S₂₂ = S₁₁ + 55
S₂₂ = 44 + 55 [S₁₁ = 44]
S₂₂ = 99
Sn = n /2 [ 2a + ( n - 1) d]
99 = (22/2)[2a + (22-1) d]
99 = (22/2)[2a + 21d]
2a + 21 d = (99 x 2)/22
2a + 21 d = 9
2 a + 21 d = 9 . …. …. (2)

On Subtracting eq (1) and (2)
2a + 10 d = 8
2 a + 21 d = 9
(-) (-) (-)
-------------------
-11 d = -1
d = 1/11

Put d = 1/11 in equation 1
2a + 10 d = 8
2 a + 10 (1/11) = 8
2 a + 10/11 = 8
2 a = 8 - (10/11)
2 a = (88 - 10)/11
2 a = 78/11
a = 78/(2x11)
a = 39/11

General form of an AP.:
a, a+d, a+2d, a+3d…….

Hence, the series is (39/11) + (40/11) + (41/11) + ...

HOPE THIS WILL HELP YOU...

Answer 2

formula for sum of n terms = n/2[2a + (n - 1)d].

now,

Sum of first 11 terms = 44

S₁₁ = 44

(11/2)[2a + (11-1) d] = 44

2a + 10 d = (44 x 2)/11

2a + 10 d = 4 x 2

2 a + 10 d = 8 ----- ( 1 )

Sum of next 11 term = 55

S₂₂ = S₁₁ + 55

S₂₂ = 44 + 55

S₂₂ = 99

(22/2)[2a + (22-1) d] = 99

2a + 21 d = (99 x 2)/22

2a + 21 d = 9

2 a + 21 d = 9 ----- ( 2 )

from ------ ( 1 ) ------ ( 2 )

we get,


-11 d = -1

d = 1/11 [ put in----( 1 )]



2 a + 10 (1/11) = 8

2 a + 10/11 = 8

2 a = 8 - (10/11)

2 a = (88 - 10)/11

2 a = 78/11

a = 78/(2x11)

a = 39/11

$\bold{\therefore}$ the series is (39/11) + (40/11) + (41/11) + ............