You are here
Home > Class 11 > Sequence and series Class 11

# Sequence and series Class 11

Chapter 9

Sequence
and series

Q. Let the sequence an be defined by a1 = 1, an = an-1 +2 for n ≥ 2.

Write first 3 terms

a1 = 1

an = an-1 +2

a2 = a2-1 +2

= a1 + 2 = 1 + 2 = 3

a3 = a3-1 +2

= a2 + 2 = 3 + 2 = 5

d = a2 – a1

If a, b and c are three consecutive terms of an A.P, 2b = a + c

an = a + (n-1)d

a2 = a + d

a3 = a + 2d

a4 = a + 3d

n = ((a_(n- ) a_1)/d) + 1

Sn = n/2(2a+(n-1)d)

= n/2(a1+an)

If a, b and c are three consecutive terms of an A.P, then b is called arithmetic mean of a and c

A.M = (a+b)/2

Q. In an A.P, the first term is 2 and sum of the first five terms is one-fourth of the next five terms. Find the 20th term

a = 2,

a1 + a2 + a3 + a4 + a5 = 1/4 (a6 + a7 + a8 + a9 + a10)

a + a + d + a + 2d + a + 3d + a+ 4d = 1/4 (a + 5d + a+ 6d + a+ 7d + a + 8d + a + 9d)

5a+10d = 1/4 (5a+35d)

4(5a+10d) = 5a + 35d

20a + 40d = 5a + 35d

20.2 + 40d = 5.2 + 35d

40 + 40d = 10 + 35d

d = -6

a20 = a + 19d

= 2 + 19.-6

= 2 – 114 = -112

Q. If the sum of n terms of an A.P is 3n2+5n and mth term is 164. Find m

an = 3n2+ 5n

a = 3.12+ 5.1 = 8

d = 2 .coefficient of n2 = 2.3 = 6

a+ (m-1)d = 164

8 + (m-1).6 = 164

8+6m-6 = 164

m=27

Q. Insert 5 numbers between 8 and 26 such that the resulting sequence is an A.P

a = 8

a+6d = 26

8+6d = 26

d = 3

11,14,17,20,23

Q. In an A.P, if mth term is n and nth term is m, m≠n, find the pth term

a + (m-1)d = n

a + (n-1)d = m

m-1-(n-1) d = n-m

(m-1-n+1)d = n-m

(m-n)d = n-m

d = (n-m)/(m-n) = (-(m-n))/(m-n) = -1

a + (m-1)d = n

a + (m-1).-1= n

a -m +1 = n

a = m+n-1

a + (p-1)d

= m+n-1 + (p-1).-1

= m+n-1-p+1

= m+n-p

Q. If the sum of first p terms of an A.P is equal to sum of first q terms. Find the sum of first p+q terms.

Sp = Sq

p/2(2a+(p-1)d)= q/2(2a+(q-1)d)

2ap+ p(p-1)d= 2aq + q(q-1)d

2ap-2aq + p(p-1)d – q(q-1)d = 0

2a(p-q) + d( p2-p-q2+q) = 0

2a(p-q) + d[(p2-q2) -p+q] = 0

2a(p-q) + d[(p+q)(p-q) –(p-q)] = 0

2a+d(p+q-1) = 0

Sp+q = (P+q)/2(2a+(p+q-1)d)

= 0

Q. The sum of first two A.Ps are in the ratio (3n+8) : (7n+15). Find the the ratio of their 12th terms.

(Sum of first n terms of first A.P)/(Sum of first n terms of second A.P)  = (3n+8)/(7n+15)

(n/2(2a_1+(n-1) d_1))/(n/2(2a_2+(n-1) d_2))  = (3n+8)/(7n+15)

((2a_1+(n-1) d_1))/((2a_2+(n-1) d_2))  = (3n+8)/(7n+15)

(a_1+ (n-1)/2  d_1)/(a_2+ (n-1)/2  d_2 ) = (3n+8)/(7n+15)               (n-1)/2 = 11, n-1 = 22, n = 23

(a_1+ (23-1)/2  d_1)/(a_2+ (23-1)/2  d_2 ) = (3.23+8)/(7.23+15)

(a_1+ 11 d_1)/(a_2+ 11 d_2 ) = (3.23+8)/(7.23+15) = 77/176 = 7/16

Geometric Progression (G.P)

3,6,12,24,…

r= a_2/a_1

an = a rn-1

a2 = ar

a3 = ar2

a4 = ar3

Sn = a((r^n-1)/(r-1))

S∞ = a/(1-r)  , |r| <1

If a, b and c are three consecutive terms of a G.P, then b is called geometric  mean of a and c

G.M = √ab

Q. Which term of the sequence 2, 2√2, 4, … is 128

a = 2

r= a_2/a_1  = (2√2)/2 = √2

an = 128

a rn-1 = 128

2. √2n-1 = 128

√2 n-1 = 64

√2 n-1 = 26 = √2 12

n-1 = 12

n = 13

Q. The sum of first three terms of a G.P is 16 and the sum of next three terms is 128. Find first term and common difference

a1 + a2 + a3 = 16

a + ar + ar2 = 16

a(1+r+r2 ) = 16

a4+a5+a6 = 128

ar3+ar4+ar5 = 128

ar3(1+r+r2) = 128

r3 = 8

r = 2

a(1+r+r2) = 16

a(1+2+22) =16

a(7) = 16

a= 16/7

Q. If the 4th , 10th and 16th terms of a G.P are x,y,z respectively. Prove that x,y,z  are in G.P

ar3 = x

ar9 = y

ar15 = z

y/x = (ar^9)/(ar^3 ) = r6

z/y = (ar^15)/(ar^9 ) = r6

x,y,z  are in G.P

Q. Find the sum to n terms of the sequence 8,88,888,…

8 + 88 + 888 + … to n terms

= 8[1 + 11 + 111 + … to n terms]

= 8/9 [9 + 99 + 999 + … to n terms]

= 8/9 [(10-1) + (100-1) + (1000-1) + … to n terms]

= 8/9 [(10 + 100 + 1000 + … to nerms)-(1 + 1+ 1+ … to n terms)]

= 8/9 [ 10((10^n-1)/(10-1)) – n]