CBSE Class 11 Mathematics Sequences And Series Worksheet Set 08

CBSE Class 11 Mathematics Worksheet - Sequences and Series (7). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. The sum of \( n \) terms of two A.P.’s are in the ratio \( (3n + 8) : (7n + 15) \). Find the ratio of their \( 12^{\text{th}} \) terms.
Answer:
\( 1^{\text{st}} \) A.P.
First term: \( a \)
Difference: \( d \)
\( 12^{\text{th}} \) term: \( a_{12} \)
Sum: \( S_n \)

\( 2^{\text{nd}} \) A.P.
First term: \( a' \)
Difference: \( d' \)
\( 12^{\text{th}} \) term: \( a'_{12} \)
Sum: \( S'_n \)

To find: \( \frac{a_{12}}{a'_{12}} \) i.e., \( \frac{a+11d}{a'+11d'} \)

Given: \( \frac{S_n}{S'_n} = \frac{3n+8}{7n+15} \)
\( \Rightarrow \frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2a'+(n-1)d']} = \frac{3n+8}{7n+15} \)
\( \Rightarrow \frac{2a+(n-1)d}{2a'+(n-1)d'} = \frac{3n+8}{7n+15} \)

Put \( n = 23 \) both the sides
\( \Rightarrow \frac{2a+22d}{2a'+22d'} = \frac{69+8}{161+15} \)
\( \Rightarrow \frac{2(a+11d)}{2(a'+11d')} = \frac{77}{176} \)
\( \Rightarrow \frac{a_{12}}{a'_{12}} = \frac{7}{16} \)

Hence, the required ratio is \( 7:16 \).

Question. The ratio of the sum of \( m \) & \( n \) terms of an A.P.’s is \( m^2 : n^2 \). Show that the ratio of the \( m^{\text{th}} \) term and \( n^{\text{th}} \) terms is \( (2m - 1) : (2n - 1) \).
Answer:
To prove: \( \frac{a_m}{a_n} = \frac{2m-1}{2n-1} \)

Given: \( \frac{S_m}{S_n} = \frac{m^2}{n^2} \)
\( \Rightarrow \frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]} = \frac{m^2}{n^2} \)
\( \Rightarrow \frac{2a+(m-1)d}{2a+(n-1)d} = \frac{m}{n} \)
\( \Rightarrow 2an + (nm-n)d = 2am + (nm-m)d \)
\( \Rightarrow 2a(n-m) + d(nm - n - nm + m) = 0 \)
\( \Rightarrow 2a(n-m) - d(n-m) = 0 \)
\( \Rightarrow (n-m)[2a-d] = 0 \)
\( \Rightarrow (2a-d) = 0 \)
\( \Rightarrow d = 2a \)

Now, \( \frac{a_m}{a_n} = \frac{a+(m-1)(2a)}{a+(n-1)(2a)} \)
\( = \frac{a+(2am-2a)}{a+(2an-2a)} \)
\( = \frac{a(1+2m-2)}{a(1+2n-2)} \)
\( \Rightarrow \frac{a_m}{a_n} = \frac{2m-1}{2n-1} \) (proved)

Question. If the sum of \( n \) terms of an A.P. is \( pn + qn^2 \). Find the common difference.
Answer:
We have, \( S_n = pn + qn^2 \).
Put \( n = 1 \), \( S_1 = p + q \)
\( \Rightarrow a_1 = p + q \qquad \{ \because S_1 = a_1 \} \)
Put \( n = 2 \), \( S_2 = 2p + 4q \)
\( \Rightarrow a_1 + a_2 = 2p + 4q \qquad \{ \because S_2 = a_1 + a_2 \} \)
\( \Rightarrow p + q + a_2 = 2p + 4q \)
\( \Rightarrow a_2 = p + 3q \)

Now, \( d = a_2 - a_1 \)
\( = (p + 3q) - (p + q) \)
\( d = 2q \).

Question. The interior angles of a polygon are in A.P. The smallest angle is \( 120^\circ \) & the common difference is \( 5^\circ \). Find the number of sides of the polygon.
Answer:
Let \( n \rightarrow \) no. of sides in the polygon
Interior angles form an A.P. with \( a = 120^\circ \), \( d = 5^\circ \), no. of term = \( n \)
Then, \( S_n = \frac{n}{2}[240 + (n-1)5] \)
\( = \frac{n}{2}[240 + 5n - 5] \)
\( S_n = \frac{n}{2}[5n + 235] \) ............ (i)

Also, sum of all interior angles in any polygon with \( n \)-sides = \( (n-2) \times 180^\circ \) ............ (ii)

Equation (i) & (ii)
\( \Rightarrow \frac{n}{2}[5n + 235] = (n-2) \times 180^\circ \)
\( \Rightarrow 5n^2 + 235n = (n-2) \times 360^\circ \)
\( \Rightarrow 5n^2 + 235n = 360n - 720 \)
\( \Rightarrow 5n^2 - 125n + 720 = 0 \)
\( \Rightarrow n^2 - 25n + 144 = 0 \)
\( \Rightarrow (n-16)(n-9) = 0 \)
\( \Rightarrow n = 16 \) or \( n = 9 \)

When \( n = 16 \),
Then, \( a_{16} = a + 15d \)
\( = 120 + 15(5) \)
\( = 195 > 180^\circ \) (not possible \( \times \) interior angle cannot \( > 180^\circ \))

When \( n = 9 \),
Then, \( a_9 = a + 8d \)
\( = 120 + 8(5) \)
\( = 160 < 180^\circ \) (possible)
\( \therefore \) no. of sides in the polygon = \( 9 \).

Question. The sum of the first \( p, q, r \) terms of an A.P. are \( a, b, c \) respectively. Show that \( \frac{a}{p}(q - r) + \frac{b}{q}(r - p) + \frac{c}{r}(p - q) = 0 \)
Answer:
Let \( A \rightarrow \) 1st term of A.P.
\( D \rightarrow \) common difference
Then \( a_p = a = \frac{p}{2}[2A + (p-1)D] \)
\( \text{(or) } \frac{a}{p} = \frac{1}{2}[2A + (p-1)D] \)
\( \Rightarrow a_q = b = \frac{q}{2}[2A + (q-1)D] \)
\( \text{(or) } \frac{b}{q} = \frac{1}{2}[2A + (q-1)D] \)
And \( a_r = c = \frac{r}{2}[2A + (r-1)D] \)
\( \text{(or) } \frac{c}{r} = \frac{1}{2}[2A + (r-1)D] \)

Now, taking L.H.S., \( \frac{a}{p}(q-r) + \frac{b}{q}(r-p) + \frac{c}{r}(p-q) \)
Putting values of \( \frac{a}{p}, \frac{b}{q}, \frac{c}{r} \) from the above equations:
\( = \frac{1}{2}[2A + (p-1)D](q-r) + \frac{1}{2}[2A + (q-1)D](r-p) + \frac{1}{2}[2A + (r-1)D](p-q) \)
\( = \frac{1}{2} \{ 2A(q-r) + (p-1)D(q-r) + 2A(r-p) + (q-1)D(r-p) + 2A(p-q) + (r-1)D(p-q) \} \)
\( = \frac{1}{2} \{ 2A[q-r+r-p+p-q] + D[p(q-r) - (q-r) + q(r-p) - (r-p) + r(p-q) - (p-q)] \} \)
\( = \frac{1}{2} \{ 2A[q-r+r-p+p-q] + D[pq - pr - q + r + qr - qp - r + p + rp - rq - p + q] \} \)
\( = \frac{1}{2} [2A(0) + D(0)] \)
\( = \frac{1}{2}(0) \)
\( = 0 = \text{R.H.S.} \)

Question. Insert 3 A.M.’s between 3 and 19.
Answer:
Here, \( a = 3, b = 19 \) & \( n = 3 \)
Let A.M.’s are \( A_1, A_2, \& A_3 \)
Now, \( d = \frac{b-a}{n+1} = \frac{19-3}{3+1} = \frac{16}{4} = 4 \)
\( A_1 = a + d = 3 + 4 = 7 \)
\( A_2 = a + 2d = 3 + 8 = 11 \)
\( A_3 = a + 3d = 3 + 12 = 15 \)
\( \dots \) required numbers are 7, 11, 15.

Question. For what value of \( n \), \( \frac{a^{n+1}+b^{n+1}}{a^n+b^n} \) is the A.M. between \( a \) & \( b \).
Answer:
We have, \( \frac{a^{n+1}+b^{n+1}}{a^n+b^n} = \text{A.M.} \)
\( \Rightarrow \frac{a^{n+1}+b^{n+1}}{a^n+b^n} = \frac{a+b}{2} \)
\( \Rightarrow 2a^{n+1} + 2b^{n+1} = (a+b)(a^n + b^n) \)
\( \Rightarrow 2a^{n+1} + 2b^{n+1} = a^{n+1} + ab^n + ba^n + b^{n+1} \)
\( \Rightarrow 2a^{n+1} - a^{n+1} + 2b^{n+1} - b^{n+1} = ab^n + ba^n \)
\( \Rightarrow a^{n+1} + b^{n+1} = ab^n + ba^n \)
\( \Rightarrow a^{n+1} - ba^n = ab^n - b^{n+1} \)
\( \Rightarrow a^n(a - b) = b^n(a - b) \)
\( \Rightarrow a^n = b^n \)
\( \Rightarrow \frac{a^n}{b^n} = 1 \)
\( \Rightarrow \left(\frac{a}{b}\right)^n = 1 \)
\( \Rightarrow \left(\frac{a}{b}\right)^n = \left(\frac{a}{b}\right)^0 \)
\( \Rightarrow n = 0 \).

Question. Between 1 and 31, \( m \) numbers are inserted so that resulting sequence is an A.P. if the ratio of the \( 7^{\text{th}} \) & \( (m - 1)^{\text{th}} \) number is \( 5:9 \). Find the value of \( m \).
Answer:
We have, \( a = 1, b = 31 \) & \( n = m \)
Now \( d = \frac{b-a}{n+1} \Rightarrow d = \frac{31-1}{m+1} = \frac{30}{m+1} \)
Given, \( \frac{A_7}{A_{m-1}} = \frac{5}{9} \)
\( \Rightarrow \frac{a+7d}{a+(m-1)d} = \frac{5}{9} \)
\( \Rightarrow \frac{1+7\left(\frac{30}{m+1}\right)}{1+(m-1)\left(\frac{30}{m+1}\right)} = \frac{5}{9} \)
\( \Rightarrow \frac{m+1+210}{m+1+30m-30} = \frac{5}{9} \)
\( \Rightarrow \frac{m+211}{31m-29} = \frac{5}{9} \) (Note: The source document has a typo listing \( 19 \) instead of \( 29 \) in the denominator of this line, but evaluates the calculation using the correct value of \( 29 \))
\( \Rightarrow 9m + 1899 = 155m - 145 \)
\( \Rightarrow 146m = 2044 \)
\( \Rightarrow m = \frac{2044}{146} = 14 \).
\( \Rightarrow m = 14 \text{ years.} \)

Question. If \( a \left( \frac{1}{b} + \frac{1}{c} \right) , b \left( \frac{1}{c} + \frac{1}{a} \right) , c \left( \frac{1}{a} + \frac{1}{b} \right) \) are in A.P. show that \( a, b, c \) are also in A.P.
Answer:
We have, \( a \left( \frac{1}{b} + \frac{1}{c} \right) , b \left( \frac{1}{c} + \frac{1}{a} \right) , c \left( \frac{1}{a} + \frac{1}{b} \right) \) are in A.P.
Adding 1 in each term:
\( \Rightarrow a \left( \frac{1}{b} + \frac{1}{c} \right) + 1, b \left( \frac{1}{c} + \frac{1}{a} \right) + 1, c \left( \frac{1}{a} + \frac{1}{b} \right) + 1 \) are also in A.P.
\( \Rightarrow a \left[ \frac{1}{b} + \frac{1}{c} + \frac{1}{a} \right], b \left[ \frac{1}{c} + \frac{1}{a} + \frac{1}{b} \right], c \left[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right] \) are in A.P.
\( \Rightarrow 2b \left[ \frac{1}{c} + \frac{1}{a} + \frac{1}{b} \right] = a \left[ \frac{1}{b} + \frac{1}{c} + \frac{1}{a} \right] + c \left[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right] \)
\( \Rightarrow 2b \left[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right] = \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) (a + c) \)
\( \Rightarrow 2b = a + c \)
\( a, b, c \) are in A.P. (proved)

Question. If the sum of three numbers in A.P. is 24 & their product is 440. Find the numbers.
Answer:
Let the numbers be \( a - d, a, a + d \)
Sum = 24
\( \dots a - d + a + a + d = 24 \)
\( \Rightarrow 3a = 24 \)
\( \Rightarrow a = 8 \)

Product = 440
\( \Rightarrow (a - d)(a)(a + d) = 440 \)
Put \( a = 8 \)
\( \Rightarrow (8 - d)(8)(8 + d) = 440 \)
\( \Rightarrow (8 - d)(8 + d) = \frac{440}{8} = 55 \)
\( \Rightarrow 64 - d^2 = 55 \)
\( \Rightarrow d^2 = 9 \)
\( \Rightarrow d = 3 \) & \( d = -3 \)

For \( a = 8 \) & \( d = 3 \):
Numbers are 11, 8, 5

\( \dots \) required numbers are 5, 8, 11 (or) 11, 8, 5.