Problems – Sequence, Series and A.P.
1.
Write
the first negative term of the sequence 20, 19 
, 18
, 17
...
Solution.
Given
sequence is 20, 19 ,
18,
17...
Hence, a
= 20, d =
– 20 = 
= 
(it is an A.P)
Let
nth term be its first negative term, i.e.
Tn
= a + (n-1) d < 0
ð
20 + (n-1) (-
)
< 0
ð 80
– 3n + 3 < 0
ð 83
– 3n < 0
ð
n > 
ð
n > 27
ð Hence,
n = 28
2.
Determine
the number of terms in A.P. 3, 7, 11 ... 407. Also, find its 11th term from the
end.
Solution.
Given
A.P. is 3, 7, 11 ... 407
Here, a =
3, d = 4, l = 407
Using the
formula, Tn = l =
a + (n-1) d, we get
407 = 3 +
(n-1) 4
ð 4n
= 408
ð n
= 102
We need 11th term from the end
Last term = 102th
Second last term = 102 – 1 = 101th
Third last term = 102 – 2 = 100th
And so on
So, 11th term from the end = (102
– 10) term = 92th term
Therefore, T92 = a + (92-1) d
= 3 + 91 x 4
= 3 + 364
= 367
OR
11th term from the end = l + (n-1)
(-d)
= 407 + (11-1) (-4)
= 407 – 40
= 367
3.
How
many numbers are there between 200 and 500, which leave remainder 7 when
divided by 9.
Solution.
The first
number lying between 200 and 500, which is divisible by 9 and leaves the
remainder 7 is 205.
The last
number lying between 200 and 500, which is divisible by 9 and leaves remainder
7 is 493
Therefore,
the numbers lying between 200 and 500, which are divisible by 9 and leave the
remainder 7 are: 205, 214, 223,…493
It is an
A.P. in which a = 205 and d = 9
Therefore,
Tn = l
= a + (n-1) d
ð 493
= 205 + (n-1) 9
ð 288
= 9n – 9
ð
= n
ð n
= 33
4.
If
in an A.P. 
, find 
Solution.
Given,
Let
the first term and common difference of A.P. be ‘A’ and ‘D’, respectively
Therefore,
ð 7A
+ 42D = 5A + 45D
ð 2A
= 3D
ð
A = 
D ….(i)
Now, 
=
(using (i))
= 
=
5.
If
the 1st 2nd and last terms of an A.P. are a, b and c, respectively then find
the sum of all terms of the A.P.
Solution.
Given
A.P. is
A, b,…c
First
term (A) = a
Common
difference (d) = b – a
Last term
i.e. nth term (An) = c
Using the
formula, An = A + (n-1) d, we
get
c = a +
(n-1) (b-a)
ð c
– a = (n-1) (b-a)
ð
n – 1 = 
ð n
= 
ð
n = 
….(i)
Now, using the formula Sn = 
(A
+ An) for sum of n terms, we get
Sn = [a
+ c]
= 
(using (i))
= 
6.
The
ratio of the sum of n terms of two A.P.'s is (7n - 1) :
(3n + 11), Find the ratio of their 10th terms.
Solution.
Let a1,
a2, be the first terms and d1, d2 be the
common differences of the two given A.P.’s. Then the sums of their n terms are
given by
Sn
= 
= 
It is
given that 
= 
ð
…(ii)
To find the ratio of the 10th
terms of the two A.P.’s, we replace n by (2 x 10 – 1) i.e. ‘19’ (i)
Replacing n by ‘19’ in (i), we get
Therefore, = 
ð
= 
= 
ð
= 
Therefore, ratio of the 10th terms
of the two A.P.’s =
7.
Find
the sum of the sequence, -1, 
, 
, 
,
…
Solution.
Given
sequence is A.P.
-1, ,
, ,
…
Where
a = -1, d = 
and l =
Using
the formula l = a + (n-1) d, we get = -1
+ (n-1) 
ð 
+
1 = 
-
ð
+ =
ð
=
ð n
= 27
Now, sum of nth terms of an
A.P. is
Sn = [2a + (n-1) d]
Therefore, S15 = 
[2(-1) + (26) ]
= [-2 +
]
= 
[
=
8.
Solve:
1+6+11+16+...+x = 148
Solution.
Clearly,
terms of the given series form an A.P. with first term a = 1 and common
difference d = 5. Let there be n terms in this series. Then, 1+6+11+16+...+x =
148
ð Sum
of n terms = 148
ð
[2a
+ (n-1) d] = 148
ð
[2
+ (n-1) 5] = 148
ð 5n2
– 3n – 296 = 0
ð (n-8)
(5n + 37) = 0
ð n
= 8 [since, n is not negative]
Now, x = nth term
ð x
= a + (n-1) d
ð x
= 1 + (8 -1) x 5 = 36 [since, a = 1, d
= 5, n = 8]
9.
If
log2, log(2n - 1) and log(2n + 3) are in A.P. Show that n
= (log5)/(log2)
Solution.
Given,
log2, log(2n - 1) and log(2n +
3) are in A.P.
Therefore,
log(2n + 3) - log(2n - 1) = log(2n
- 1) – log 2
ð log
(
= log (
Taking antilog, we get
ð 
=
ð 22n
– 4.2 n – 5 = 0
ð y2
– 4y - 5 = 0 where y = 2n
ð y
= 5 and y = -1 (reject)
Hence, 2n = 5
Or nlog2 = log 5 or n = 
Hence Proved
10.
If
a,b,c are in A.P. show that
following are also in A.P.
i)
, 
, 
ii)
b
+ c, c + a, a + b
Solution.
i)
Given a, b, c are in A.P.
ð
, 
, 
are
also in A.P. [On dividing each by abc]
ð
, , are
also in A.P.
ii)
b + c, c + a, a + b will be in
A.P.
if
(c + a) – (b + c) = (a + b) – ( c + a)
i.e.
if a – b = b – c
i.e.
if 2b = a + c
i.e.
if a, b, c are in A.P.
Thus, a, b, c are in A.P.
ð b
+ c, c + a, a + b are in A.P.