Chapter 9 Sequences and Series Ex 9.2 Solutions
Question - 1 : - Find the sum of odd integers from 1 to 2001.
Answer - 1 : -
The odd integers from1 to 2001 are 1, 3, 5, …1999, 2001.
It clearly forms asequence in A.P.
Where, the firstterm, a = 1
Commondifference, d = 2
Now,
a + (n -1)d = 2001
1 + (n-1)(2) = 2001
2n – 2 = 2000
2n = 2000 + 2 = 2002
n = 1001
We know,
Sn =n/2 [2a + (n-1)d]
Therefore, the sum ofodd numbers from 1 to 2001 is 1002001.
Question - 2 : - Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
Answer - 2 : -
The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.
It clearly forms a sequence in A.P.
Where, the first term, a = 105
Common difference, d = 5
Now,
a + (n -1)d = 995
105 + (n – 1)(5) = 995
105 + 5n – 5 = 995
5n = 995 – 105 + 5 = 895
n = 895/5
n = 179
We know,
Sn = n/2 [2a + (n-1)d]
Therefore, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.
Question - 3 : - In an A.P, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is –112.
Answer - 3 : -
Given,
The first term (a) ofan A.P = 2
Let’s assume d bethe common difference of the A.P.
So, the A.P. will be2, 2 + d, 2 + 2d, 2 + 3d, …
Then,
Sum of first fiveterms = 10 + 10d
Sum of next five terms= 10 + 35d
From the question, wehave
10 + 10d = ¼ (10 +35d)
40 + 40d = 10 + 35d
30 = -5d
d = -6
a20 =a + (20 – 1)d = 2 + (19) (-6) = 2 – 114 = -112
Therefore, the 20th termof the A.P. is –112.
Question - 4 : - How many terms of the A.P. -6, -11/2, -5, …. are needed to give the sum –25?
Answer - 4 : -
Let’s consider the sum of n terms of the given A.P. as –25.
We known that,
Sn = n/2 [2a + (n-1)d]
where n = number of terms, a = first term, and d = common difference
So here, a = –6
d = -11/2 + 6 = (-11 + 12)/ 2 = 1/2
Thus, we have
Question - 5 : - In an A.P., if pth term is 1/q and qth term is 1/p, prove that the sum of first pq terms is ½ (pq + 1) where p ≠ q.
Answer - 5 : -
Question - 6 : - If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term
Answer - 6 : -
Given A.P.,
25, 22, 19, …
Here,
First term, a = 25 and
Common difference, d =22 – 25 = -3
Also given, sum ofcertain number of terms of the A.P. is 116
The number of terms ben
So, we have
Sn =n/2 [2a + (n-1)d] = 116
116 = n/2 [2(25) +(n-1)(-3)]
116 x 2 = n [50 – 3n +3]
232 = n [53 – 3n]
232 = 53n – 3n2
3n2 –53n + 232 = 0
3n2 –24n – 29n+ 232 = 0
3n(n – 8) – 29(n – 8)= 0
(3n – 29) (n – 8) = 0
Hence,
n = 29/3 or n = 8
As n can only be anintegral value, n = 8
Thus, 8th termis the last term of the A.P.
a8 =25 + (8 – 1)(-3)
= 25 – 21
= 4
Question - 7 : - Find the sum to n terms of the A.P., whose kth term is 5k + 1.
Answer - 7 : -
Given, the kth termof the A.P. is 5k + 1.
kth term = ak = a +(k – 1)d
And,
a + (k – 1)d = 5k +1
a + kd – d =5k + 1
On comparing thecoefficient of k, we get d = 5
a – d = 1
a – 5 = 1
⇒ a =6
Question - 8 : - If the sum of n terms of an A.P. is (pn + qn2),where p and q are constants, find the commondifference.
Answer - 8 : -
We know that,
Sn =n/2 [2a + (n-1)d]
From the question wehave,
On comparing thecoefficients of n2 on both sides, we get
d/2 = q
Hence, d =2q
Therefore, the commondifference of the A.P. is 2q.
Question - 9 : - The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6. Find the ratio of their 18th terms.
Answer - 9 : -
Let a1, a2,and d1, d2 be the firstterms and the common difference of the first and second arithmetic progressionrespectively.
Then, from thequestion we have
Question - 10 : - If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.
Answer - 10 : -
Let’s take a and d tobe the first term and the common difference of the A.P. respectively.
Then, it given that
Therefore, the sum of(p + q) terms of the A.P. is 0.