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]
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