11th NCERT/CBSE Sequences and series Exercise 9.2 Questions 12
Hi

Question (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 forms an A.P.
Here, first term, a = 1
Common difference, d = 2
Here, a + (n-1)d = 2001
⇒ 1 +(n-1)(2) = 2001
⇒ 2n-2 = 2000
⇒ n = 1001
${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$ \[{S_n} = \frac{{1001}}{2}\left[ {2 \times 1 + \left( {1001 - 1} \right) \times 2} \right]\] \[{S_n} = \frac{{1001}}{2}\left[ {2 + 1000 \times 2} \right]\] \[{S_n} = \frac{{1001}}{2} \times 2002\] \[{S_n} = 1001 \times 1001 = 1002001\] 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

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 = 995
⇒ 105 +(n-1)5 = 995
⇒ (n-1)5 = 995 - 105 = 890
⇒ n -1 = 178
⇒ n = 179
\[ \therefore {S_n} = \frac{{179}}{2}\left[ {2\left( {105} \right) + \left( {179 - 1} \right)\left( 5 \right)} \right]\] \[ = 179\left[ {105 + \left( {89} \right)5} \right]\] \[ = 179\left[ {105 + 445} \right]\] \[ = 98450\] 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 terms is one-fourthh 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,
$10 + 10d = \frac{1}{4}\left( {10 + 35d} \right)$
$ \Rightarrow 40 + 40d = 10 + 35d$
$ \Rightarrow 30 = - 5d$
$ \Rightarrow d = - 6$
∴ a20 = a + (20-1)d
a20 = 2 + (19)(-6) = 2 -144 = -112
Thus, the 20th term of the A.P. is –112.

Question (4)

How many terms of the A.P. -6, $ - \frac{{11}}{2}$, -5 , ... are needed to give sum -25

Solution

Let the sum of n terms of the given A.P. be –25.
It is known that, ${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$ , where n = number of terms
a = first term and d = common difference
Here, a = - 6
$d = - \frac{{11}}{2} + 6 = \frac{{ - 11 + 12}}{2} = \frac{1}{2}$
Therefore, we obtain
$ - 25 = \frac{n}{2}\left[ {2 \times \left( { - 6} \right) + \left( {n - 1} \right)\left( {\frac{1}{2}} \right)} \right]$
\[ \Rightarrow - 50 = n\left[ { - 12 + \frac{n}{2} - \frac{1}{2}} \right]\] \[ \Rightarrow - 50 = n\left[ { - \frac{{25}}{2} + \frac{n}{2}} \right]\] \[ \Rightarrow - 100 = n\left( { - 25 + n} \right)\] \[ \Rightarrow {n^2} - 25n + 100 = 0\] \[ \Rightarrow {n^2} - 5n - 20n + 100 = 0\] \[ \Rightarrow n\left( {n - 5} \right) - 20\left( {n - 5} \right) = 0\] \[ \Rightarrow n = 20 \; or \; 5\]

Question (5)

In an A.P., if pth term is $\frac{1}{q}$ and qth term is $\frac{1}{p}$, prove that the sum of first pq terms is $\frac{1}{2}\left( {pq + 1} \right)$, 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,
\[{p^{th}}\;term = {a_p} = a + \left( {p - 1} \right)d = \frac{1}{q} - - - (1)\] \[ {q^{th}}\;term= {a_q} = a + \left( {q - 1} \right)d = \frac{1}{p} - - - (2)\] Subtracting (2) from (1), we obtain \[\left( {p - 1} \right)d - \left( {q - 1} \right)d = \frac{1}{q} - \frac{1}{p}\] \[ \Rightarrow \left( {p - 1 - q + 1} \right)d = \frac{{p - q}}{{pq}}\] \[ \Rightarrow \left( {p - q} \right)d = \frac{{p - q}}{{pq}}\] \[ \Rightarrow d = \frac{1}{{pq}}\] Putting the value of d in (1), we obtain \[a + \left( {p - 1} \right)\frac{1}{{pq}} = \frac{1}{q}\] \[ \Rightarrow a = \frac{1}{q} - \frac{1}{q} + \frac{1}{{pq}} = \frac{1}{{pq}}\] \[\therefore {S_{pq}} = \frac{{pq}}{2}\left[ {2a + \left( {pq - 1} \right)d} \right]\] \[ = \frac{{pq}}{2}\left[ {\frac{2}{{pq}} + \left( {pq - 1} \right)\frac{1}{{pq}}} \right]\] \[ = 1 + \frac{1}{2}\left( {pq - 1} \right)\] \[ = \frac{1}{2}pq + 1 - \frac{1}{2} = \frac{1}{2}pq + \frac{1}{2}\] \[ = \frac{1}{2}\left( {pq + 1} \right)\] Thus, the sum of first pq terms of the A.P. is $ = \frac{1}{2}\left( {pq + 1} \right)$

Question (6)

`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 given A.P. be 116.
\[{S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\] Here, a = 25 and d = 22 – 25 = – 3 \[{S_n} = \frac{n}{2}\left[ {2 \times 25 + \left( {n - 1} \right)\left( { - 3} \right)} \right]\] \[ \Rightarrow 116 = \frac{n}{2}\left[ {50 - 3n + 3} \right]\] \[ \Rightarrow 232 = n\left( {53 - 3n} \right) = 53n - 3{n^2}\] \[ \Rightarrow 3{n^2} - 53n + 232 = 0\] \[ \Rightarrow 3{n^2} - 24n - 29n + 232 = 0\] \[ \Rightarrow 3n\left( {n - 8} \right) - 29\left( {n - 8} \right) = 0\] \[ \Rightarrow \left( {n - 8} \right)\left( {3n - 29} \right) = 0\] \[ \Rightarrow n = 8 \; or \;n = \frac{{29}}{3}\] However, n cannot be equal to $\frac{{29}}{3}$ . 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

Solution

It is given that the kth term of the A.P. is 5k + 1.
kth 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 \[{S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\] \[{S_n} = \frac{n}{2}\left[ {2\left( 6 \right) + \left( {n - 1} \right)\left( 5 \right)} \right]\] \[{S_n} = \frac{n}{2}\left[ {12 + 5n - 5} \right]\] \[{S_n} = \frac{n}{2}\left( {5n + 7} \right)\]

Question (8)

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

Solution

It is known that, ${S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
According to the given condition,
\[\frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] = pn + q{n^2}\] \[ \Rightarrow \frac{n}{2}\left[ {2a + nd - d} \right] = pn + q{n^2}\] \[ \Rightarrow na + {n^2}\frac{d}{2} - n \cdot \frac{d}{2} = pn + q{n^2}\] Comparing the coefficients of n2 on both sides, we obtain $\frac{d}{2} = q$
∴ d = 2q 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 there 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, \[\frac{{sum\;of\;n\;terms\;of\;first\;A.P.}}{{sum\;of\;n\;terms\;of \;second A.P.}} = \frac{{5n + 4}}{{9n + 6}}\] \[ \Rightarrow \frac{{\frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right){d_1}} \right]}}{{\frac{n}{2}\left[ {2{a_2} + \left( {n - 1} \right){d_2}} \right]}} = \frac{{5n + 4}}{{9n + 6}}\] \[ \Rightarrow \frac{{2{a_1} + \left( {n - 1} \right)d_1}}{{2{a_2} + \left( {n - 1} \right){d_2}}} = \frac{{5n + 4}}{{9n + 6}} - - - (1)\] Substituting n = 35 in (1), we obtain \[\frac{{2{a_1} + 34{d_1}}}{{2{a_2} + 34{d_2}}} = \frac{{5\left( {35} \right) + 4}}{{9\left( {35} \right) + 6}}\] \[ \Rightarrow \frac{{{a_1} + 17{d_1}}}{{{a_2} + 17{d_2}}} = \frac{{179}}{{321}} - - - (2)\] \[\frac{{{{18}^{th}}termoffirstA.P.}}{{{{18}^{th}}termof\sec ondA.P}} = \frac{{{a_1} + 17{d_1}}}{{{a_2} + 17{d_2}}} ----(3)\] From (2) and (3), we obtain \[\frac{{{{18}^{th}}term\;of\;first\;A.P.}}{{{{18}^{th}}termof\sec ondA.P}} = \frac{{179}}{{321}}\] 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.

Solution

Let a and d be the first term and the common difference of the A.P. respectively. Here,
\[{S_p} = \frac{p}{2}\left[ {2a + \left( {p - 1} \right)d} \right]\] \[{S_q} = \frac{p}{2}\left[ {2a + \left( {q - 1} \right)d} \right]\] According to the given condition, \[\frac{p}{2}\left[ {2a + \left( {p - 1} \right)d} \right] = \frac{q}{2}\left[ {2a + \left( {q - 1} \right)d} \right]\] \[ \Rightarrow p\left[ {2a + \left( {p - 1} \right)d} \right] = q\left[ {2a + \left( {q - 1} \right)d} \right]\] \[ \Rightarrow 2ap + pd\left( {p - 1} \right) = 2aq + qd\left( {q - 1} \right)\] \[ \Rightarrow 2a\left( {p - q} \right) + d\left[ {p\left( {p - 1} \right) - q\left( {p - 1} \right)} \right] = 0\] \[ \Rightarrow 2a\left( {p - q} \right) + d\left[ {{p^2} - p - {q^2} + q} \right] = 0\] \[ \Rightarrow 2a\left( {p - q} \right) + d\left[ {\left( {p - q} \right)\left( {p + q} \right) - \left( {p - q} \right)} \right] = 0\] \[ \Rightarrow 2a\left( {p - q} \right) + d\left[ {\left( {p - q} \right)\left( {p + q - 1} \right)} \right] = 0\] \[ \Rightarrow 2a + d\left( {p + q - 1} \right) = 0\] \[ \Rightarrow d = \frac{{ - 2a}}{{p + q - 1}} - - - (1)\] \[{S_{p + q}} = \frac{{p + q}}{2}\left[ {2a + \left( {p + q - 1} \right) \cdot d} \right]\] From (1)
\[{S_{p + q}} = \frac{{p + q}}{2}\left[ {2a + \left( {p + q - 1} \right) \cdot \left( {\frac{{ - 2a}}{{p + q - 1}}} \right)} \right]\] \[{S_{p + q}} = \frac{{p + q}}{2}\left[ {2a - a} \right]\] \[{S_{p + q}} = 0\] Thus, the sum of the first (p + q) terms of the A.P. is 0.

Question (11)

Sum of p, q and r terms of an A.P. are a, b and c, respectively. Prove that
\[\frac{a}{p}\left( {q - r} \right) + \frac{b}{q}\left( {r - p} \right) + \frac{c}{r}\left( {p - q} \right) = 0\]

Solution

Sp = a, Sq = b, Sr = c
Let 1st term of A.P. be 'A' and difference be 'd'
Sp=a
\[\frac{p}{2}\left[ {2A + \left( {p - 1} \right)d} \right] = a\] \[A + \left( {p - 1} \right)\frac{d}{2} = \frac{a}{p} - - - (1)\] Sq=b
\[\frac{q}{2}\left[ {2A + \left( {q - 1} \right)d} \right] = b\] \[A + \left( {q - 1} \right)\frac{d}{2} = \frac{b}{q} - - - (2)\] Sr = c
\[\frac{r}{2}\left[ {2A + \left( {r - 1} \right)d} \right] = c\] \[A + \left( {r - 1} \right)\frac{d}{2} = \frac{c}{r} - - - (3)\] Replaciing value from (1),(2) and (3) in LHS \[LHS = \frac{a}{p}\left( {q - r} \right) + \frac{b}{q}\left( {r - p} \right) + \frac{c}{r}\left( {p - q} \right)\] \[ = \left[ {A + \left( {p - 1} \right)\frac{d}{2}} \right]\left( {q - r} \right) + \left[ {A + \left( {q - 1} \right)\frac{d}{2}} \right]\left( {r - p} \right) + \left[ {A + \left( {r - 1} \right)\frac{d}{2}} \right]\left( {p - q} \right)\] \[ = A\left( {q - r} \right) + \frac{d}{2}\left( {p - 1} \right)\left( {q - r} \right) + \left( {r - p} \right)A + \left( {q - 1} \right)\left( {r - p} \right)\frac{d}{2} + A\left( {p - q} \right) + \left( {p - q} \right)\left( {r - 1} \right)\frac{d}{2}\] \[ = A\left[ {q - r + r - p + p - q} \right] + \frac{d}{2}\left[ {\left( {p - 1} \right)\left( {q - r} \right) + \left( {q - 1} \right)\left( {r - p} \right) + \left( {p - q} \right)\left( {r - 1} \right)} \right]\] \[ = 0 + \frac{d}{2}\left[ {pq - pr - q + r + qr - pq - r + p + pr - p - qr + q} \right]\] \[ = \frac{d}{2}\left( 0 \right) = 0 = RHS\]

Question (12)

The ratio of the sums of m and n terms of an A.P. is m2:n2. Show that the ratio of mth and nth terms is (2m -1) : (2n-1)

Solution

\[\frac{{{S_m}}}{{{S_n}}} = \frac{{{m^2}}}{{{n^2}}}\] Let first term be a and diifference is 'd' for A.P.
$\frac{{\frac{m}{2}\left[ {2a + \left( {m - 1} \right)d} \right]}}{{\frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]}} = \frac{{{m^2}}}{{{n^2}}}$
$\frac{{2a + \left( {m - 1} \right)d}}{{2a + \left( {n - 1} \right)d}} = \frac{m}{n}$
2an + (m-1)nd = 2am + (n-1)md
(m-1)nd-(n-1)md = 2am -2an
d[mn - n - mn +m] = 2a(m-n)
d(m - n) = 2a (m-n)
d = 2a
$\frac{{{a_m}}}{{{a_n}}} = \frac{{a + \left( {m - 1} \right)d}}{{a + \left( {n - 1} \right)d}}$
$\frac{{{a_m}}}{{{a_n}}} = \frac{{a + \left( {m - 1} \right)2a}}{{a + \left( {n - 1} \right)2a}}$
$\frac{{{a_m}}}{{{a_n}}} = \frac{{a\left[ {1 + 2m - 2} \right]}}{{a\left[ {1 + 2n - 2} \right]}}$
$\frac{{{a_m}}}{{{a_n}}} = \frac{{2m - 1}}{{2n - 1}}$

Question (13)

If the sum of n terms of A.P. is 3n2 + 5n and its mth term is 164, find the value of m

Solution

3n2 + 5n, am = 164, m =?
S1 = a1 = 8
S2 = 3(2)2 + 5(2) = 12 +10 = 22
a2 = S2 - S1
a2 = 22 - 8 = 14
d = a2 - a1 = 14 -8 = 6
a = 8 , d = 6
am = 164
a + (m-1)d = 164
8 + (m-1) 6 = 164
$\left( {m - 1} \right) = \frac{{164 - 8}}{6}$
$\left( {m - 1} \right) = \frac{{156}}{6}$
$m - 1 = 26$
$m = 27$

Question (14)

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

Solution

a=8, b =26, n =5
$d = \frac{{b - a}}{{n + 1}} = \frac{{26 - 8}}{6} = \frac{{18}}{6}$
d = 3
The terms are 8+3 =11, 11+3 = 14 and so on
The inserted A.P. = 11, 14, 17, 20, 23

Question (15)

If $\frac{{{a^n} + {b^n}}}{{{a^{n - 1}} + {b^{n - 1}}}}$ is the A.M. between a nand b, then find the value of n

Solution

A.M. of a and b $ = \frac{{a + b}}{2}$
According to the given condition,
$\frac{{a + b}}{2} = \frac{{{a^n} + {b^n}}}{{{a^{n - 1}} + {b^{n - 1}}}}$
⇒ (a+b)(an-1 + bn-1) = 2(an + bn)
⇒ an + abn-1 + ban-1 + nn = 2an+2bn
⇒ abn-1 + an-1b = an + bn
⇒ abn-1 - bn = an - an-1b
⇒ bn-1( a - b) = an-1 (a-b)
⇒ bn-1 = an-1
$ \Rightarrow {\left( {\frac{a}{b}} \right)^{n - 1}} = 1 = {\left( {\frac{a}{b}} \right)^0}$
⇒ n - 1 = 0
⇒ n = 1

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 number is 5:9. Find the value of m

Solution

a = 1, b 31, n = m
\[d = \frac{{b - a}}{{n + 1}} = \frac{{31 - 1}}{{m + 1}} = \frac{{30}}{{m + 1}}\] 7th inserted term = 8th term of A.P. starting from 1
(m-1)th inserted term = mth term of A.P. starting from 1
$\frac{{{a_8}}}{{{a_m}}} = \frac{5}{9} \Rightarrow \frac{{a + 7d}}{{a + \left( {m - 1} \right)d}}$
Replace a =1
$\frac{{1 + 7d}}{{1 + \left( {m - 1} \right)d}} = \frac{5}{9}$
⇒ 9 + 63d = 5 +5(m-1)d
(63 - 5m +5)d = 5 -9
(68 - 5m)d = -4
Replace 'd'
$\left( {68 - 5m} \right)\frac{{30}}{{m + 1}} = - 4$
2040- 150m = -4m -4
2044 = 146m
m =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?

Solution

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.

Solution

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).
${S_n} = {180^o}\left( {n - 2} \right)$
\[ \Rightarrow \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] = {180^o}\left( {n - 2} \right)\] \[ \Rightarrow \frac{n}{2}\left[ {240 + \left( {n - 1} \right)5} \right] = 180\left( {n - 2} \right)\] \[ \Rightarrow n\left[ {240 + \left( {n - 1} \right)5} \right] = 360\left( {n - 2} \right)\] \[ \Rightarrow 240n + 5{n^2} - 5n = 360n - 720\] \[ \Rightarrow 5{n^2} + 235n - 360n + 720 = 0\] \[ \Rightarrow 5{n^2} - 125n + 720 = 0\] \[ \Rightarrow {n^2} - 25n + 144 = 0\] \[ \Rightarrow {n^2} - 16n - 9n + 144 = 0\] \[ \Rightarrow n\left( {n - 16} \right) - 9\left( {n - 16} \right) = 0\] \[ \Rightarrow \left( {n - 9} \right)\left( {n - 16} \right) = 0\] \[ \Rightarrow n = 9 \;or\;16\]
Exercise 9.1⇐
⇒Exercise 9.3