Sequence and Series JEE Mathematics Worksheets Set 01

Subjective Questions

Question. The third term of an A.P. is 18, and the seventh term is 30 ; find the sum of 17 terms.
Answer: Let \( a \) be the first term and \( d \) be the common difference.
\( a + 2d = 18 \)
\( a + 6d = 30 \)
By subtracting the first equation from the second,
\( \implies 4d = 12 \)
\( \implies d = 3 \text{ and } a = 12 \)
The sum of 17 terms is \( S_{17} = \frac{17}{2} [2(12) + (17 - 1)3] \)
\( \implies S_{17} = \frac{17}{2} [24 + 48] = 17 [12 + 24] = 17 \times 36 = 612 \).

Question. Find the number of integers between 100 & 1000 that are
(i) divisible by 7
(ii) not divisible by 7
Answer: (i) The integers between 100 and 1000 divisible by 7 are 105, 112, ..., 994.
Using \( \ell = a + (n - 1)d \),
\( 994 = 105 + (n - 1)7 \)
\( \implies 889 = (n - 1)7 \)
\( \implies n - 1 = 127 \)
\( \implies n = 128 \).
(ii) Total integers between 100 and 1000 are \( 999 - 100 = 899 \).
Integers not divisible by 7 = Total integers - integers divisible by 7
\( \implies 899 - 128 = 771 \).

Question. Find the sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7.
Answer: The integers are of the form \( 16k + 7 \).
The first integer is \( 16(6) + 7 = 103 \).
The last integer is \( 16(49) + 7 = 791 \).
Number of terms \( n = 49 - 6 + 1 = 44 \).
Sum \( S_{44} = \frac{44}{2} [103 + 791] \)
\( \implies S_{44} = 22 \times 894 = 19668 \).

Question. The sum of three numbers in A.P. is 27, and their product is 504, find them.
Answer: Let the numbers be \( a - d, a, a + d \).
Sum: \( (a - d) + a + (a + d) = 27 \)
\( \implies 3a = 27 \)
\( \implies a = 9 \)
Product: \( (a - d) \cdot a \cdot (a + d) = 504 \)
\( \implies 9(81 - d^2) = 504 \)
\( \implies 81 - d^2 = 56 \)
\( \implies d^2 = 25 \)
\( \implies d = \pm 5 \).
The numbers are 4, 9, 14 or 14, 9, 4.

Question. If a, b, c are in A.P., then show that
(i) \( a^2 (b + c), b^2 (c + a), c^2 (a + b) \) are also in A.P.
(ii) \( b + c - a, c + a - b, a + b - c \) are in A.P.
Answer: (i) Since \( a, b, c \) are in A.P., we have \( 2b = a + c \).
Consider \( a^2(b + c) + abc, b^2(c + a) + abc, c^2(a + b) + abc \).
\( \implies a(ab + ac + bc), b(bc + ba + ac), c(ca + cb + ab) \).
Dividing by \( (ab + bc + ca) \), we get \( a, b, c \), which are in A.P. Therefore, the original terms are in A.P.
(ii) Since \( a, b, c \) are in A.P., \( -2a, -2b, -2c \) are in A.P.
Adding \( a + b + c \) to each term,
\( \implies (a + b + c - 2a), (a + b + c - 2b), (a + b + c - 2c) \) are in A.P.
\( \implies b + c - a, c + a - b, a + b - c \) are in A.P.

Question. The continued product of three numbers in G.P. is 216, and the sum of the products of them in pairs is 156, find the numbers.
Answer: Let the numbers be \( a/r, a, ar \).
Product: \( \frac{a}{r} \cdot a \cdot ar = 216 \)
\( \implies a^3 = 216 \)
\( \implies a = 6 \)
Sum of products in pairs: \( \frac{a}{r} \cdot a + a \cdot ar + ar \cdot \frac{a}{r} = 156 \)
\( \implies a^2 (\frac{1}{r} + r + 1) = 156 \)
\( \implies 36 (\frac{1 + r^2 + r}{r}) = 156 \)
\( \implies 3(r^2 + r + 1) = 13r \)
\( \implies 3r^2 - 10r + 3 = 0 \)
\( \implies (r - 3)(3r - 1) = 0 \)
\( \implies r = 3 \text{ or } r = 1/3 \).
The numbers are 2, 6, 18 or 18, 6, 2.

Question. If the \( p^{th}, q^{th}, r^{th} \) terms of a G.P. be a, b, c respectively, prove that \( a^{q - r} b^{r - p} c^{p - q} = 1 \).
Answer: Let \( A \) be the first term and \( R \) be the common ratio.
\( a = AR^{p-1}, b = AR^{q-1}, c = AR^{r-1} \)
\( a^{q-r} b^{r-p} c^{p-q} = (AR^{p-1})^{q-r} (AR^{q-1})^{r-p} (AR^{r-1})^{p-q} \)
\( \implies A^{(q-r)+(r-p)+(p-q)} R^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)} \)
\( \implies A^0 R^{pq-pr-q+r+qr-pq-r+p+rp-rq-p+q} = A^0 R^0 = 1 \).

Question. The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 & the third is increased by 1, we obtain three consecutive terms of a G.P., find the numbers.
Answer: Let the numbers be \( a - d, a, a + d \).
Sum: \( 3a = 21 \)
\( \implies a = 7 \).
The numbers in A.P. are \( 7 - d, 7, 7 + d \).
According to the condition, \( 7 - d, 7 - 1, 7 + d + 1 \) are in G.P.
\( \implies 7 - d, 6, 8 + d \) are in G.P.
\( \implies 6^2 = (7 - d)(8 + d) \)
\( \implies 36 = 56 - d - d^2 \)
\( \implies d^2 + d - 20 = 0 \)
\( \implies (d + 5)(d - 4) = 0 \)
\( \implies d = -5 \text{ or } d = 4 \).
If \( d = -5 \), numbers are 12, 7, 2.
If \( d = 4 \), numbers are 3, 7, 11.

Question. The sum of infinite number of terms of G.P. is 4 and the sum of their cubes is 192. Find the series.
Answer: Let the series be \( a, ar, ar^2, \dots \)
Sum \( S_\infty = \frac{a}{1 - r} = 4 \)
\( \implies a = 4(1 - r) \) ...(1)
Sum of cubes \( S_{cubes} = \frac{a^3}{1 - r^3} = 192 \)
\( \implies \frac{[4(1 - r)]^3}{1 - r^3} = 192 \)
\( \implies \frac{64(1 - r)^3}{(1 - r)(1 + r + r^2)} = 192 \)
\( \implies \frac{(1 - r)^2}{1 + r + r^2} = 3 \)
\( \implies 1 - 2r + r^2 = 3 + 3r + 3r^2 \)
\( \implies 2r^2 + 5r + 2 = 0 \)
\( \implies (2r + 1)(r + 2) = 0 \)
\( \implies r = -1/2 \) (as \( |r| < 1 \)).
From (1), \( a = 4(1 + 1/2) = 6 \).
The series is 6, -3, 3/2, -3/4, ...

Question. Sum the following series
(i) \( 1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + \dots \) to n terms.
(ii) \( 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + \dots \) to infinity.
Answer: (i) It is an A.G.P. with common ratio \( r = 1/2 \).
Sum \( S = 4 - \frac{n + 2}{2^{n - 1}} \).
(ii) \( T_n = \frac{2^n - 1}{4^{n - 1}} = \frac{2^n}{2^{2n - 2}} - \frac{1}{2^{2n - 2}} = \frac{1}{2^{n - 2}} - \frac{1}{2^{2n - 2}} \).
\( S = \sum_{n = 1}^\infty (\frac{1}{2^{n - 2}} - \frac{1}{2^{2n - 2}}) = (2 + 1 + \frac{1}{2} + \dots) - (1 + \frac{1}{4} + \frac{1}{16} + \dots) \)
\( \implies S = \frac{2}{1 - 1/2} - \frac{1}{1 - 1/4} = 4 - \frac{4}{3} = \frac{8}{3} \).

Question. Find the sum of n terms of the series the \( r^{th} \) term of which is \( (2r + 1)2^r \).
Answer: \( S = \sum_{r = 1}^n (2r + 1)2^r \)
\( S = 3 \cdot 2 + 5 \cdot 2^2 + 7 \cdot 2^3 + \dots + (2n + 1)2^n \)
\( 2S = 3 \cdot 2^2 + 5 \cdot 2^3 + \dots + (2n - 1)2^n + (2n + 1)2^{n + 1} \)
Subtracting equations:
\( -S = 6 + 2(2^2 + 2^3 + \dots + 2^n) - (2n + 1)2^{n + 1} \)
\( \implies -S = 6 + 2^3 \frac{2^{n - 1} - 1}{2 - 1} - (2n + 1)2^{n + 1} \)
\( \implies S = (2n + 1)2^{n + 1} - 2^{n + 2} + 2 = 2^{n + 1}(2n - 1) + 2 \).

Question. Find the \( 4^{th} \) term of an H.P. whose \( 7^{th} \) term is \( \frac{1}{20} \) and \( 13^{th} \) term is \( \frac{1}{38} \).
Answer: In corresponding A.P., \( T_7 = 20 \text{ and } T_{13} = 38 \).
\( a + 6d = 20 \)
\( a + 12d = 38 \)
Solving, \( 6d = 18 \)
\( \implies d = 3 \text{ and } a = 2 \).
For A.P., \( T_4 = a + 3d = 2 + 3(3) = 11 \).
In H.P., \( T_4 = 1/11 \).

Question. The arithmetic mean of two numbers is 6 and their geometric mean G and harmonic mean H satisfy the relation \( G^2 + 3 H = 48 \). Find the two numbers.
Answer: \( A = \frac{a + b}{2} = 6 \)
\( \implies a + b = 12 \).
Also, \( G^2 = AB = 6H \).
Given \( G^2 + 3H = 48 \)
\( \implies 6H + 3H = 48 \)
\( \implies 9H = 48 \)
\( \implies H = 16/3 \).
\( G^2 = 6(16/3) = 32 \).
\( \implies ab = 32 \).
Solving \( a + b = 12 \text{ and } ab = 32 \), we get \( a = 4, b = 8 \) or vice versa.

Question. Using the relation A.M. \( \ge \) G.M. prove that
(i) \( \tan \theta + \cot \theta \ge 2 \); if \( 0 < \theta < \frac{\pi}{2} \)
(ii) \( (x^2y + y^2z + z^2x) (xy^2 + yz^2 + zx^2) > 9x^2y^2z^2 \). Where x,y,z are different real no.
(iii) \( (a + b) \cdot (b + c) \cdot (c + a) \ge 8abc \); if a, b, c are positive real numbers.
Answer: (i) \( \frac{\tan \theta + \cot \theta}{2} \ge \sqrt{\tan \theta \cdot \cot \theta} = 1 \)
\( \implies \tan \theta + \cot \theta \ge 2 \).
(ii) For different \( x, y, z \), A.M. > G.M.
\( \frac{x^2y + y^2z + z^2x}{3} > (x^2y \cdot y^2z \cdot z^2x)^{1/3} = (x^3y^3z^3)^{1/3} = xyz \)
\( \implies x^2y + y^2z + z^2x > 3xyz \).
Similarly, \( xy^2 + yz^2 + zx^2 > 3xyz \).
Multiplying gives \( (x^2y + y^2z + z^2x)(xy^2 + yz^2 + zx^2) > 9x^2y^2z^2 \).
(iii) \( \frac{a + b}{2} \ge \sqrt{ab}, \frac{b + c}{2} \ge \sqrt{bc}, \frac{c + a}{2} \ge \sqrt{ca} \).
Multiplying these gives \( \frac{(a + b)(b + c)(c + a)}{8} \ge \sqrt{a^2b^2c^2} = abc \)
\( \implies (a + b)(b + c)(c + a) \ge 8abc \).

Question. Find the sum of the n terms of the series whose nth term is
(i) \( n(n + 2) \)
(ii) \( 3^n - 2^n \)
Answer: (i) \( S_n = \sum (n^2 + 2n) = \frac{n(n + 1)(2n + 1)}{6} + \frac{2n(n + 1)}{2} \)
\( \implies S_n = \frac{n(n + 1)(2n + 7)}{6} \).
(ii) \( S_n = \sum (3^n - 2^n) = \frac{3(3^n - 1)}{2} - \frac{2(2^n - 1)}{1} \)
\( \implies S_n = \frac{3^{n + 1} - 3}{2} - 2^{n + 1} + 2 = \frac{3^{n + 1} + 1}{2} - 2^{n + 1} \).

Question. The sum of the first ten terms of an AP is 155 & the sum of first two terms of a GP is 9. The first term of the AP is equal to the common ratio of the GP & the first term of the GP is equal to the common difference of the AP. Find the two progressions.
Answer: Let AP: \( a, a + d, \dots \) and GP: \( A, AR, \dots \).
Given \( S_{10} = \frac{10}{2} [2a + 9d] = 155 \)
\( \implies 2a + 9d = 31 \).
Also, \( A + AR = 9 \). Given \( a = R \) and \( A = d \).
\( \implies d(1 + a) = 9 \)
\( \implies d = \frac{9}{1 + a} \).
Substituting in the first equation:
\( 2a + \frac{81}{1 + a} = 31 \)
\( \implies 2a^2 - 29a + 50 = 0 \)
\( \implies (2a - 25)(a - 2) = 0 \)
\( \implies a = 2 \text{ or } a = 25/2 \).
If \( a = 2, d = 3 \). AP: 2, 5, 8, ... and GP: 3, 6, 12, ...
If \( a = 25/2, d = 2/3 \). AP: 25/2, 79/6, 83/6, ... and GP: 2/3, 25/3, 625/6, ...

Question. Find the sum in the \( n^{th} \) group of sequence,
(i) 1, (2, 3); (4, 5, 6, 7); (8, 9,........, 15) ; ...............
(ii) (1), (2, 3, 4), (5, 6, 7, 8, 9), .........
Answer: (i) In \( n^{th} \) group, terms are \( (2^{n - 1}, 2^{n - 1} + 1, \dots, 2^n - 1) \). There are \( 2^{n - 1} \) terms.
\( S_n = \frac{2^{n - 1}}{2} [2(2^{n - 1}) + (2^{n - 1} - 1) \cdot 1] = 2^{n - 2} [2^n + 2^{n - 1} - 1] \).
(ii) \( n^{th} \) group has \( (2n - 1) \) terms ending with \( n^2 \).
The sum of all terms up to \( n^{th} \) group is \( \sum_{i = 1}^{n^2} i = \frac{n^2(n^2 + 1)}{2} \).
Sum of \( n^{th} \) group \( S_n = n^3 + (n - 1)^3 \).

Question. Find the sum of the series \( \frac{5}{13} + \frac{55}{(13)^2} + \frac{555}{(13)^3} + \frac{5555}{(13)^4} \dots \) up to \( \infty \).
Answer: \( S = \frac{5}{9} (\frac{9}{13} + \frac{99}{13^2} + \frac{999}{13^3} + \dots) \)
\( \implies S = \frac{5}{9} [(\frac{10}{13} + \frac{100}{13^2} + \dots) - (\frac{1}{13} + \frac{1}{13^2} + \dots)] \)
\( \implies S = \frac{5}{9} [\frac{10/13}{1 - 10/13} - \frac{1/13}{1 - 1/13}] = \frac{5}{9} [\frac{10}{3} - \frac{1}{12}] \)
\( \implies S = \frac{5}{9} \times \frac{39}{12} = \frac{65}{36} \).

Question. If \( 0 < x < \pi \) and the expression \( \exp \{ (1 + |\cos x| + \cos^2 x + |\cos^3 x| + \cos^4 x + \dots \text{ upto } \infty) \log_e 4 \} \) satisfies the quadratic equation \( y^2 - 20y + 64 = 0 \) then find the value of x.
Answer: \( y^2 - 20y + 64 = 0 \)
\( \implies (y - 16)(y - 4) = 0 \)
\( \implies y = 16 \text{ or } y = 4 \).
The exponent is \( \frac{1}{1 - |\cos x|} \log_e 4 \).
So \( 4^{\frac{1}{1 - |\cos x|}} = 4^2 \text{ or } 4^1 \).
\( \implies \frac{1}{1 - |\cos x|} = 2 \text{ or } 1 \).
If 2, \( |\cos x| = 1/2 \implies x = \pi/3, 2\pi/3 \).
If 1, \( |\cos x| = 0 \implies x = \pi/2 \).

Question. In a circle of radius R a square is inscribed, then a circle is inscribed in the square, a new square in the circle and so on for n times. Find the limit of the sum of areas of all the circle and the limit of the sum of areas of all the squares as \( n \to \infty \).
Answer: Radius decreases by \( 1/\sqrt{2} \) and side by \( 1/\sqrt{2} \).
Sum of areas of circles = \( \pi R^2 [1 + 1/2 + 1/4 + \dots] = 2\pi R^2 \).
Side of first square is \( \sqrt{2}R \). Area is \( 2R^2 \).
Sum of areas of squares = \( 2R^2 [1 + 1/2 + 1/4 + \dots] = 4R^2 \).

Question. If a, b, c are sides of triangle then prove that
(i) \( b^2c^2 + c^2a^2 + a^2b^2 \ge abc(a + b + c) \)
(ii) \( (a + b + c)^3 > 27(a + b - c)(c + a - b)(b + c - a) \).
Answer: (i) Using A.M. \( \ge \) G.M.: \( \frac{b^2c^2 + c^2a^2}{2} \ge \sqrt{b^2c^2 \cdot c^2a^2} = abc^2 \).
Adding similar relations gives \( b^2c^2 + c^2a^2 + a^2b^2 \ge abc^2 + a^2bc + ab^2c = abc(a + b + c) \).
(ii) Using A.M. > G.M. for \( (a + b - c), (c + a - b), (b + c - a) \):
Sum is \( a + b + c \).
\( \implies \frac{a + b + c}{3} > [(a + b - c)(c + a - b)(b + c - a)]^{1/3} \)
\( \implies (a + b + c)^3 > 27(a + b - c)(c + a - b)(b + c - a) \).

Question. Sum the following series to n terms and to infinity
(i) \( \sum_{r=1}^n r(r + 1)(r + 2)(r + 3) \)
(ii) \( \frac{n}{1.2.3} + \frac{n-1}{2.3.4} + \dots + \frac{1}{n(n+1)(n+2)} \).
Answer: (i) \( T_r = \frac{1}{5} [r(r + 1)(r + 2)(r + 3)(r + 4) - (r - 1)r(r + 1)(r + 2)(r + 3)] \).
\( S_n = \frac{n(n + 1)(n + 2)(n + 3)(n + 4)}{5} \).
(ii) \( S_n = \sum_{r = 1}^n \frac{n - r + 1}{r(r + 1)(r + 2)} = \frac{n(n + 1)}{4(n + 2)} \).

Question. Sum of the series to n terms and to infinity :
\( 1^2 - \frac{2^2}{5} + \frac{3^2}{5^2} - \frac{4^2}{5^3} + \frac{5^2}{5^4} - \frac{6^2}{5^5} + \dots \infty \).
Answer: The sum to infinity is \( S = \frac{25}{54} \).

Question. In an A.P. of which 'a' is the Ist term, if the sum of the Ist 'p' terms is equal to zero, show that the sum of the next 'q' terms is \( -\frac{a(p + q)q}{p - 1} \).
Answer: \( S_p = \frac{p}{2} [2a + (p - 1)d] = 0 \)
\( \implies d = -\frac{2a}{p - 1} \).
Sum of next \( q \) terms is \( S_{p + q} - S_p = S_{p + q} = \frac{p + q}{2} [2a + (p + q - 1)d] \)
\( \implies S_{p + q} = \frac{p + q}{2} [2a + (p + q - 1)(-\frac{2a}{p - 1})] \)
\( \implies S_{p + q} = a(p + q) [1 - \frac{p + q - 1}{p - 1}] = a(p + q) [\frac{p - 1 - p - q + 1}{p - 1}] = -\frac{a(p + q)q}{p - 1} \).

Question. The number of terms in an A.P. is even; the sum of the odd terms is 24, sum of the even terms 30, and the last term exceeds the first by \( 10.5 \); find the number of terms.
Answer: Let there be \( 2n \) terms.
Sum of odd terms: \( n[a + (n - 1)d] = 24 \).
Sum of even terms: \( n[a + d + (n - 1)d] = 30 \).
Subtracting equations: \( nd = 6 \).
Also, \( T_{2n} - T_1 = (2n - 1)d = 10.5 \).
\( 2nd - d = 10.5 \)
\( \implies 2(6) - d = 10.5 \)
\( \implies d = 1.5 \).
\( n = 6/1.5 = 4 \). Total terms \( 2n = 8 \).

Question. A man arranges to pay off debit of Rs. 3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid he dies leaving a third of the debt unpaid. Find the value of the first instalment.
Answer: \( S_{40} = 20[2a + 39d] = 3600 \)
\( \implies 2a + 39d = 180 \). ...(1)
\( S_{30} = 3600 - 1/3(3600) = 2400 \).
\( \implies 15[2a + 29d] = 2400 \)
\( \implies 2a + 29d = 160 \). ...(2)
Subtracting (2) from (1): \( 10d = 20 \implies d = 2 \).
From (2), \( 2a + 58 = 160 \implies 2a = 102 \implies a = 51 \).
The first instalment is 51 Rs.

Question. If the \( p^{th}, q^{th} \) and \( r^{th} \) terms of an A.P. are a, b, c respectively, show that
\( (q - r)a + (r - p) b + (p - q) c = 0 \).
Answer: \( a = A + (p - 1)d, b = A + (q - 1)d, c = A + (r - 1)d \).
\( b - c = (q - r)d \implies q - r = \frac{b - c}{d} \).
Similarly, \( r - p = \frac{c - a}{d} \) and \( p - q = \frac{a - b}{d} \).
Substituting in the expression: \( a \frac{b - c}{d} + b \frac{c - a}{d} + c \frac{a - b}{d} = \frac{ab - ac + bc - ba + ca - cb}{d} = 0 \).

Question. If b is the harmonic mean between a and c prove that
\( \frac{1}{b - a} + \frac{1}{b - c} = \frac{1}{a} + \frac{1}{c} \).
Answer: \( b = \frac{2ac}{a + c} \).
L.H.S. \( \frac{1}{b - a} + \frac{1}{b - c} = \frac{2b - (a + c)}{b^2 - b(a + c) + ac} \)
Substituting \( b(a + c) = 2ac \):
\( \implies \frac{2b - (a + c)}{b^2 - 2ac + ac} = \frac{2b - (a + c)}{b^2 - ac} \)
Since \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in A.P., \( \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \text{ or } 2ac = b(a + c) \). The relation is identical to \( \frac{1}{a} + \frac{1}{c} \).

Question. Circles are inscribed in the acute angle \( \alpha \) so that every neighbouring circles touch each other. If the radius of the first circle is R then find the sum of the radii of the first n circles in terms of R and \( \alpha \).
Answer: Let radii be \( R, R_2, R_3, \dots \).
Using geometry, \( R_2 = R \frac{1 + \sin(\alpha/2)}{1 - \sin(\alpha/2)} \).
It is a G.P. with ratio \( r = \frac{1 + \sin(\alpha/2)}{1 - \sin(\alpha/2)} \).
Sum \( S_n = R \frac{r^n - 1}{r - 1} = \frac{R}{2 \sin(\alpha/2)} [(\frac{1 + \sin(\alpha/2)}{1 - \sin(\alpha/2)})^n - 1] \).

Question. The first term of arithmetic progression is 1 and the sum of the first nine terms equal to 369. The first and the ninth term of a geometric progression coincide with the first and the ninth term of the arithmetic progression. Find the seventh term of the geometric progression.
Answer: For AP, \( S_9 = \frac{9}{2} [2(1) + 8d] = 369 \)
\( \implies 2 + 8d = 82 \implies d = 10 \).
\( T_9 = 1 + 8(10) = 81 \).
For GP, \( A = 1 \) and \( T_9 = AR^8 = 81 \)
\( \implies R^8 = 3^4 \implies R = \sqrt{3} \).
\( T_7 = AR^6 = 1 \cdot (\sqrt{3})^6 = 3^3 = 27 \).