DPP for JEE: Daily Practice Problems- Sequences & Series (Solutions)

 Page 2


or ... (i)
or 
or ... (ii)
Dividing (i) by (ii), we get
Applying componendo and dividendo,
We get 
2. (a) Let the first set of number be a – d, a, a + d.
Then a – d + a + a + d = 15 a = 5
The second set of numbers will be
b – (d – 1), b,  b + (d – 1). Again,
Hence, the sets of numbers are
5 – d,  5,  5 + d and 6 – d, 5, 4 + d.
Further, from the given condition
? The two sets are 3, 5, 7 and 4, 5, 6 or
21, 5, –11 and 22,  5,  –12.
? Ratio of their smallest term is 
3. (b) The general term is
Page 3


or ... (i)
or 
or ... (ii)
Dividing (i) by (ii), we get
Applying componendo and dividendo,
We get 
2. (a) Let the first set of number be a – d, a, a + d.
Then a – d + a + a + d = 15 a = 5
The second set of numbers will be
b – (d – 1), b,  b + (d – 1). Again,
Hence, the sets of numbers are
5 – d,  5,  5 + d and 6 – d, 5, 4 + d.
Further, from the given condition
? The two sets are 3, 5, 7 and 4, 5, 6 or
21, 5, –11 and 22,  5,  –12.
? Ratio of their smallest term is 
3. (b) The general term is
where
?
4. (d) Let, S
n
  = 2 + 5 +14 + 41 + .......+ x
n
5. (a) If a, ar, ar
2
, ar
3
 ......... are in G.P., then
sum of infinite G.P.= a + ar + ..... +  
where 'a' is the first term and 'r' is the common ratio of G.P.
Given 
This is a GP, with common ratio 'a'.
Page 4


or ... (i)
or 
or ... (ii)
Dividing (i) by (ii), we get
Applying componendo and dividendo,
We get 
2. (a) Let the first set of number be a – d, a, a + d.
Then a – d + a + a + d = 15 a = 5
The second set of numbers will be
b – (d – 1), b,  b + (d – 1). Again,
Hence, the sets of numbers are
5 – d,  5,  5 + d and 6 – d, 5, 4 + d.
Further, from the given condition
? The two sets are 3, 5, 7 and 4, 5, 6 or
21, 5, –11 and 22,  5,  –12.
? Ratio of their smallest term is 
3. (b) The general term is
where
?
4. (d) Let, S
n
  = 2 + 5 +14 + 41 + .......+ x
n
5. (a) If a, ar, ar
2
, ar
3
 ......... are in G.P., then
sum of infinite G.P.= a + ar + ..... +  
where 'a' is the first term and 'r' is the common ratio of G.P.
Given 
This is a GP, with common ratio 'a'.
Again, This is also a G.P., with common ratio 'b'.
Now, consider 
which is again a GP with common ratio 'ab'.
6. (d) Let , 
7. (a) We have  
? 
Page 5


or ... (i)
or 
or ... (ii)
Dividing (i) by (ii), we get
Applying componendo and dividendo,
We get 
2. (a) Let the first set of number be a – d, a, a + d.
Then a – d + a + a + d = 15 a = 5
The second set of numbers will be
b – (d – 1), b,  b + (d – 1). Again,
Hence, the sets of numbers are
5 – d,  5,  5 + d and 6 – d, 5, 4 + d.
Further, from the given condition
? The two sets are 3, 5, 7 and 4, 5, 6 or
21, 5, –11 and 22,  5,  –12.
? Ratio of their smallest term is 
3. (b) The general term is
where
?
4. (d) Let, S
n
  = 2 + 5 +14 + 41 + .......+ x
n
5. (a) If a, ar, ar
2
, ar
3
 ......... are in G.P., then
sum of infinite G.P.= a + ar + ..... +  
where 'a' is the first term and 'r' is the common ratio of G.P.
Given 
This is a GP, with common ratio 'a'.
Again, This is also a G.P., with common ratio 'b'.
Now, consider 
which is again a GP with common ratio 'ab'.
6. (d) Let , 
7. (a) We have  
? 
So, 
8. (c) If d be the common difference, then
m = a
4
 – a
7
 + a
10
 = a
4
 – a
7
 + a
7
? 
+ 3d = a
7
S
13
 = 
= 
9. (c) Given that a, b, c are in G.P.
So, … (i)
… (ii)
… (iii)
Now 
10. (b) If x > 1, y > 1, z > 1 are in G.P.
Then 
x > 1, y > 1, z > 1,
? y
2
 = xz ...(i)
Taking log on both sides of equ. (i), we get
2 log y = log x + log z
? 2 + 2 log y = 2 + log x + log z
? 2 (1 + log y) = (1 + log x) + (1 + log z)
Clearly, (1 + log x), (1 + log y) (1 + log z) are in A.P.