SaraNextGen.Com
Updated By SaraNextGen
On March 11, 2024, 11:35 AM

Exercise 9.2 - Chapter 9 Sequences & Series class 11 ncert solutions Maths - SaraNextGen [2024]


Question 1:

Find the sum of odd integers from 1 to 2001.

Answer:

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

This sequence forms an A.P.

Here, first term, a = 1

Common difference, d = 2

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4276/chapter%209_html_m1b7090ef.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4276/chapter%209_html_m3198cc78.gif

Thus, the sum of odd 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:

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4277/chapter%209_html_m74540e02.gif

Thus, 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:

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,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4278/chapter%209_html_75558220.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4278/chapter%209_html_m7aca428f.gif

Thus, the 20th term of the A.P. is –112.

Question 4:

How many terms of the A.P. https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4279/chapter%209_html_2f27a2eb.gifare needed to give the sum –25?

Answer:

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

It is known that, https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4279/chapter%209_html_35403b1e.gif, where n = number of terms, a = first term, and d = common difference

Here, a = –6

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4279/chapter%209_html_71974034.gif

Therefore, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4279/chapter%209_html_m2d7643d1.gif

Question 5:

In an A.P., if pth term is https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4280/chapter%209_html_6ef86e0.gifand qth term is https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4280/chapter%209_html_3379ed60.gif, prove that the sum of first pq terms is https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4280/chapter%209_html_m627ca353.gif

Answer:

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

∴ According to the given information,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4280/chapter%209_html_m7be101d8.gif

Subtracting (2) from (1), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4280/chapter%209_html_m33bdac1f.gif

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4280/chapter%209_html_43146cc1.gif

Thus, the sum of first pq terms of the A.P. is https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4280/chapter%209_html_m6413c914.gif.

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:

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

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4281/chapter%209_html_35403b1e.gif

Here, a = 25 and d = 22 – 25 = – 3

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4281/chapter%209_html_5d5aff93.gif

However, n cannot be equal to https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4281/chapter%209_html_m25dabe0e.gif. Therefore, n = 8

∴ a8 = 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.

Question 7:

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

Answer:

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

kth term = ak = + (k – 1)d

∴ + (k – 1)d = 5k + 1

a + kd – d = 5k + 1

Comparing the coefficient of k, we obtain d = 5

– = 1

⇒ a – 5 = 1

⇒ a = 6

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4282/chapter%209_html_358b6f.gif

Question 8:

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

Answer:

It is known that,https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/5602/Chapter%209_html_5530a2c9.gif

According to the given condition,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/5602/Chapter%209_html_3f292060.gif

Comparing the coefficients of n2 on both sides, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/5602/Chapter%209_html_m653efd96.gif

∴ d = 2 q

Thus, the common difference 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:

Let a1a2, and d1dbe the first terms and the common difference of the first and second arithmetic progression respectively.

According to the given condition,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4284/chapter%209_html_m74a7f081.gif

Substituting n = 35 in (1), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4284/chapter%209_html_m5cd65b4f.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4284/chapter%209_html_431a68eb.gif

From (2) and (3), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4284/chapter%209_html_m54aee168.gif

Thus, the ratio of 18th term of both the A.P.s is 179: 321.

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:

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

Here,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4285/chapter%209_html_73539b78.gif

According to the given condition,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4285/chapter%209_html_m47831821.gif

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

Question 11:

Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.

Prove that https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4286/chapter%209_html_331411fc.gif

Answer:

Let a1 and d be the first term and the common difference of the A.P. respectively.

According to the given information,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4286/chapter%209_html_m6447ad8e.gif

Subtracting (2) from (1), we obtain

p – 1d – q – 1d = 2ap – 2bq⇒dp – 1 – q + 1 = 2aq – 2bppq⇒dp – q = 2aq – 2bppq⇒d = 2aq – bppqp – q                …….4

Subtracting (3) from (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4286/chapter%209_html_5efe6e47.gif

Equating both the values of d obtained in (4) and (5), we obtain

aq – bppqp – q = br – qcqrq – r⇒aq – bppp – q = br – qcrq – r⇒rq – raq – bp = pp – qbr – qc⇒raq – bpq – r = pbr – qcp – q⇒aqr – bprq – r = bpr – cpqp – q

Dividing both sides by pqr, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4286/chapter%209_html_6346d0de.gif

Thus, the given result is proved.

Question 12:

The ratio of the sums of m and n terms of an A.P. is m2n2. Show that the ratio of mth and nth term is (2m – 1): (2n – 1).

Answer:

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

According to the given condition,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/5603/Chapter%209_html_6102d645.gif

Putting m = 2m – 1 and = 2n – 1 in (1), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/5603/Chapter%209_html_296d7fc2.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/5603/Chapter%209_html_m3852a091.gif

From (2) and (3), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/5603/Chapter%209_html_m3bccf8cf.gif

Thus, the given result is proved.

Question 13:

If the sum of n terms of an A.P. is https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4288/chapter%209_html_m54618244.gifand its mth term is 164, find the value of m.

Answer:

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

am = a + (m – 1)d = 164 … (1)

Sum of n terms,https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4288/chapter%209_html_35403b1e.gif

Here,

n2 2a + nd – d = 3n2 + 5n⇒na + d2n2 – d2n = 3n2 + 5n⇒d2n2 + a – d2n = 3n2 + 5n

Comparing the coefficient of n2 on both sides, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4288/chapter%209_html_3954c3c0.gif

Comparing the coefficient of n on both sides, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4288/chapter%209_html_m3efc4042.gif

Therefore, from (1), we obtain

8 + (m – 1) 6 = 164

⇒ (m – 1) 6 = 164 – 8 = 156

⇒ – 1 = 26

⇒ m = 27

Thus, the value of m is 27.

Question 14:

Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Answer:

Let A1, A2, A3, A4, and A5 be five numbers between 8 and 26 such that

8, A1, A2, A3, A4, A5, 26 is an A.P.

Here, = 8, = 26, n = 7

Therefore, 26 = 8 + (7 – 1) d

⇒ 6d = 26 – 8 = 18

⇒ = 3

A1 = a + d = 8 + 3 = 11

A2 = a + 2d = 8 + 2 × 3 = 8 + 6 = 14

A3 = a + 3d = 8 + 3 × 3 = 8 + 9 = 17

A4 = a + 4= 8 + 4 × 3 = 8 + 12 = 20

A5 = a + 5d = 8 + 5 × 3 = 8 + 15 = 23

Thus, the required five numbers between 8 and 26 are 11, 14, 17, 20, and 23.

Question 15:

If https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4292/chapter%209_html_679c4d0e.gifis the A.M. between a and b, then find the value of n.

Answer:

A.M. of a and b https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4292/chapter%209_html_m2aa0e2f6.gif

According to the given condition,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4292/chapter%209_html_111af2da.gif

Question 16:

Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7th and (m – 1)th numbers is 5:9. Find the value of m.

Answer:

Let A1, A2, … Am be m numbers such that 1, A1, A2, … Am, 31 is an A.P.

Here, a = 1, b = 31, n = m + 2

∴ 31 = 1 + (m + 2 – 1) (d)

⇒ 30 = (m + 1) d

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4293/chapter%209_html_m1c6fc479.gif

A1 = a + d

A2 = a + 2d

A3 = a + 3d …

∴ A7 = a + 7d

Am–1 = a + (m – 1) d

According to the given condition,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4293/chapter%209_html_1e266084.gif

Thus, the value of m is 14.

Question 17:

A man starts repaying a loan as first installment of Rs. 100. If he increases the installment by Rs 5 every month, what amount he will pay in the 30th installment?

Answer:

The first installment of the loan is Rs 100.

The second installment of the loan is Rs 105 and so on.

The amount that the man repays every month forms an A.P.

The A.P. is 100, 105, 110, …

First term, a = 100

Common difference, d = 5

A30 = a + (30 – 1)d

= 100 + (29) (5)

= 100 + 145

= 245

Thus, the amount to be paid in the 30th installment is Rs 245.

Question 18:

The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.

Answer:

The angles of the polygon will form an A.P. with common difference d as 5° and first term a as 120°.

It is known that the sum of all angles of a polygon with n sides is 180° (n – 2).

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4295/chapter%209_html_m341d3551.gif

Also Read : Exercise-9.3-Chapter-9-Sequences-&-Series-class-11-ncert-solutions-Maths

SaraNextGen