# Sequences and Series – exercise 9.2 [1 – 10]- Class XI

1. Find the sum of odd integers from 1 to 2001

Solution

The odd integers from 1 to 2001 are 1, 3, 5 … 1999,2001

This sequence from an A.P.

Here first term is a = 1

Common difference d = 2

Then, a + (n – 1)d = 2001

1 + (n – 1)2 = 2001

n = 1001

Sn = n/2 [2a + (n – 1)d]

Sn = 1001/2 [2×1 + (1001 – 1)x2]

= 1001/2 x [2 + 100×2]

= 1001 x 1001

= 1002001

Thus the sum of odd numbers from 1 to 2001 is 1002001

1. Find the sum of all natural numbers lying between 100 to 1000 which are multiples of 5.

Solution:

The natural numbers lying between 100 and 1000 which are multiples of 5 are 105, 110 … 995

Here a = 105 and d = 5

a + (n – 1)d = 2001

105 + (n-1)5 = 995

(n-1)5 = 995 – 105 = 890

n –  1 = 178

n = 179

Sn = n/2 [2a + (n – 1)d]

Sn = 179/2 [2×105 + (179 – 1)x5]

= 179[105 + 89×5]

= 179 x 550

= 98450

Thus the sum of all natural numbers lying between 100 and 10000 which are multiples of 5 is 98450

1. 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

Solution:

First term = 2

Let d be the common difference of the A.P.

Therefore the A.P is 2, 2 + d, 2 + 2d, 2 + 3d…

Sum of first five terms = 10 + 10d

Sum of next five terms = 10 + 35d

According to the given condition

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

40 + 40d = 10 + 35d

30 = -5d

d = -6

a20 = a + (20-1)d = 2+(19)(-6) = 2 -114 = -112

Thus the 20th term of the is -112

1. How many terms o the A.P -6, –11/2, -5… are needed to give the sum -25?

Solution:

Let the sum of n terms of the given A.P. be -25

It is known that

Sn = n/2 [2a + (n – 1)d]

Here a = -6

d = –11/2 + 6 = -11+12/2 = 1/2

Therefore, we obtain

-25 = n/2 [ 2x(-6)+(n-1)(1/2)]

-50 = n[-12+ n/21/2]

-50 = n[- 25/2 + n/2]

-100 = n(-25+n)

n2 – 25n + 100 = 0

n2 – 5n – 20n + 100 = 0

n(n-5) – 20(n – 5) = 0

n = 20 or 5

1. 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 1/2(pq + 1), where p≠q

Solution:

It is known that the general term of an A.P. is an = a + (n – 1)d

According to the given information

pth term = ap = a +(p – 1)d = 1/q  ———————(1)

qth term = aq = a +(q – 1)d = 1/p  ———————(2)

Subtracting (1) from (1), we obtain

(p – 1)d – (q-1)d = 1/q1/p

(p-1-q+1)d = p-q/pq

d = 1/pq

Putting the value of d in (1), we obtain

a+(p-1)1/pq = 1/q

a = 1/q  – 1/q + 1/pq  = 1/pq

Spq = pq/2 [2a +(pq – 1)d]

= pq/2 [2/pq + (pq – 1)1/pq]

= 1 + 1/2(pq – 1)

= 1/2pq + 1 – 1/2 = 1/2 pq + 1/2

= 1/2(pq + 1)

Thus the sum of first pq terms of the A.P. is 1/2(pq + 1)

1. If the sum of a certain number of terms of the A.P 25, 22, 19,… is 116, find the last term.

Solution:

Let the sum of n terms of the guven A.P. be the 116

Sn = n/2 [2a + (n – 1)d]

Here  a = 25 and d = 22 – 25 = -3

Sn = n/2 [2 x 25 + (n – 1)(-3)]

116 = n/2 [50 – 3n + 3]

232 = n(53 – 3n) = 53n – 3n2

3n2 – 53n + 232 = 0

3n2 – 24n – 29n + 232 = 0

(n – 8)(3n – 29) = 0

n = 8 or n = 29/3

However n cannot be equal to 29/3. Therefore n = 8

Thus, a= last term = a + (n – 1)d = 25 + (8 – 1)(-3)

= 25 + 7(-3) = 25 – 21

= 4

Thus, the last term of the A.P. is 4

1. Find the sum to n terms of the A.P., whose kth term is 5k + 1.

Solution:

It is given that the k th term of the A.P. is 5k + 1

k th term = ak=a+(k – 1)d

a + (k – 1)d = 5k + 1

a + kd – d = 5k + 1

Comparing the coefficient of k, we obtain d = 5, a – d = 1

a – 5 = 1

a = 6

Sn = n/2[2a + (n – 1)d]

= n/2[2(6) + (n – 1)5]

= n/2[12 + 5n – 5]

= n/2(5n + 7)

1. If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants find the common difference.

Solution:

It is known that Sn = n/2[2a + (n – 1)d]

According to the given condition,

n/2[2a + (n – 1)d] = pn + qn2

n/2[2a + nd – d] = pn + qn2

na + n2 .d/2 – n.d/2 = pn + qn2

Comparing the coeffients of n2 on both sides, we obtain

d/2 = q

d = 2q

Thus the common difference of A.P. is 2q

1. The sum of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n +6. Find the ratio of their 18th terms

Solution:

Let a1, a2 and d­1, d2 be the first terms and the common difference of the first and second arithmetic progression respectively.

According to the given condition,

Sum of n terms of first A.P./ Sum of n terms of second A.P = 5n+4/9n+6

Subtracting n = 35 in (1), we obtain

1. If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then ind the sum of the first (p + q) terms.

Solution:

Let a and b be the first term and the common difference of the A.P. respectively.

Sp = p/2[2a +(p-1)d]

Sq = q/2[2a +(q-1)d]

According to the given condition

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

p[a +(p-1)d] = q[a +(q-1)d]

2ap + pd(p – 1) = 2aq + qd(q – 1)

2a(p – q) + d[p2-p-q2+q]=0

2a(p – q)+d[(p-q)(p+q)-(p-q)]=0

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

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

d = -2a/p+q-1

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

= p+q/2[2a+(-2a/p+q-1 )(p+q-1)]

= p+q/2[2a – 2a]

=0

Thus the sum of the first (p+q) terms of the A.P. is 0