Sequence and Series Class 11 OP Malhotra Exe-14G ISC Maths Solutions Ch-14 Solutions. In this article you would learn about Special Types of Problems. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.
Sequence and Series Class 11 OP Malhotra Exe-14G ISC Maths Solutions Ch-14
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-14 | Sequence and Series |
| Writer | O.P. Malhotra |
| Exe-14(G) | Special Types of Problems |
Exercise- 14G
Sequence and Series Class 11 OP Malhotra Exe-14G Solution.
Que-1: Find three numbers in G.P. whose sum is 19 and product is 216 .
Sol: Let the three numbers in G.P are a/r, a, ar and their product = 216
⇒ (a/r) × a × ar = 216 ⇒ a³ = 6³ ⇒ a = 6
Also given their sum is 19 .
∴ (a/r) + a + ar = 19 ⇒ (6/r) + 6 + 6r = 19
⇒ 6 + 6r + 6r² = 19r
⇒ 6 r² – 13r + 6 = 0
⇒ r = {13±√(169−144)}/12 = (13±5)/12
⇒ r = 3/2, 2/3
When a = 6 and r = 3/2
Then required numbers are {6/(3/2)}, 6, 6 × 3/2
i.e. 4. 6. 9
when r = 6, r = 2/3
Then required numbers are {6/(2/3)}, 6, 6 × 2/3
i.e. 9, 6, 4.
Que-2: The sum of the first terms of a G.P. is (13/12) and their product is -1. Find the G.P.
Sol: Let the required G.P be, a, ar, ar² …….
it is given that a + ar + ar² = 13/12 …(1)
Also (a) (ar) (ar²) = -1 ⇒ (ar)³ = -1
⇒ ar = – 1 ⇒ a = –1/r …(2)
∴ from (1) and (2); we have
(–1/r) (–1/r) × r – (1/r) × r² = 13/12
⇒ – (1/r) – 1 – r = 13/12
⇒ – 1 – r – r² = 13r/12
⇒ – 12 – 12r – 12r² = 13r
⇒ 12r² + 25r + 12 = 0
∴ r = {−25±√(625−576)}/24 = (−25±7)/24
⇒ r = –32/24, –18/24 i.e. –4/3, –3/4
Thus required G.P be;
3/4, 3/4 × (−4/3), 3/4 (−4/3)² …..
i.e. 3/4, -1, 4/3, …..
when r = –3/4
∴ from (2); a = 4/3
Thus required G.P be;
4/3, 4/3 × (−3/4), 4/3 (−3/4)² ……
i.e. 4/3, -1, 3/4, …….
Que-3: The product of first three terms of a G.P. is 1000 . If we add 6 to its second term and 7 to its third term, the resulting three terms form an A.P. Find the terms of the G.P.
Sol: Let the required G.P be a, ar, ar2, ……
it is given that a, ar, ar2 = 1000
⇒ (ar)3 =103 ⇒ ar = 10
⇒ a = 10/r…(1)
it is given that by adding 6 to second term and 7 to third term then the resulting terms are in A.P.
i.e. a, ar + 6, ar2 + 7 forms A.P.
⇒ 2(ar + 6) = a + ar2 + 7
⇒ 2(10 + 6) = (10/r) + (10/r) . r2 + 7
⇒ 32 = (10/r)+ 10r + 7
⇒ 10r2 + 7r + 10 = 32r
⇒ 10r2 – 25r + 10 = 0
⇒ 2r2 – 5r + 2 = 0
⇒ (r – 2)(2r – 1) = 0 ⇒ r = 2, 1/2
when r = 2 ∴ from (1); a = 10/5 = 5
and infinite G.P be, 5, 10, 20, …….
when r = 1/2
∴ from (1); a = 10/(1/2) = 20
and infinite G.P be, 20, 20 × 1/2, 20 × (1/2)², …..
i.e. 20, 10, 5, …..
Que-4: The lengths of three unequal edges of a rectangular solid block are in G.P. If the volume of the block is 216 cm³ and the total surface area is 252 cm². Find the length of the longest edge.
Sol: The lengths of three unequal edges of a rectangular solid block are in G.P.
Let the lengths be A/r A, Ar
⇒ Volume = A3 = 216
⇒ A = 6
and Surface area = 2 [(6/r).6 + 6.6r + (6/r).6r] = 252
⇒ 72 {(1/r)+r+1} = 252
⇒ 2(r2 + r + 1) = 7r
⇒ 2r2 – 5r + 2 = 0
⇒ r = 2 or 1/2
So longest length = 6 × 2 = 12
Que-5: If a, b, c are in G.P show that the following are also in G.P
(i) 1/a, 1/b, 1/c
(ii) 1/a², 1/b², 1/c²
(iii) a2, b2, c2
(iv) b2c2, c2a2, a2b2
Sol: (i) Given a, b, c are in G.P
∴ b² = ac …(1)
⇒ 1/ac = 1/b²
⇒ (1/b)² = (1/a) × (1/c)
⇒ 1/a, 1/b, 1/c are in G.P
(ii) On squaring eqn. (1); we have b^4 = a²c²
⇒ 1/(a²c²) = 1/b^4
⇒ (1/a²) × (1/c²) = (1/b²)²
⇒ 1/a², 1/b², 1/c² are in G.P
(iii) On squaring eqn. (1) ; we have
b4 = (ac)2 = a2 c2 ⇒ (b2)2 = a2 × c2
Thus, a2, b2 and c2 are in G.P.
(iv) On squaring eqn. (1); we iave
b4 = a2 c2
Again squaring ; we have
⇒ (b4)2 = (a2c2)2
⇒ b4 . b4 = (a2c2)2
⇒ b4a2c2 = (a2 c2)2
⇒ (a2b2) (b2c2) = (a2c2)2
Thus b2c2, a2c2 and a2 b2 are in G.P.
Que-6: If a, b, c, d are in G.P., show that the following are also in G.P.
(i) a + b, b + c, c + d
(ii) a2 + b2, b2 + c2, c2 + d2
Sol: Given a, b, c and d are in G.P.
∴ b/a = c/b = d/c = r (say)
⇒ b = ar ; c = br = ar2; d = cr = ar3
(i) Here, (b + c)2 = (ar + ar2)2
= a2r2(1 + r)2
and (a + b)(c + d) = (a + ar) (ar2 + ar3)
= a(1 + r) ar2(1 + r)
= a2r2(1 + r)2
∴ (b + c)2 = (a + b)(c + d)
Thus a + b, b + c and c + d are in G.P.
(ii) Here, (b2 + c2)2 = (a2r2 + a2r4)2
= (a2r2)2[1 + r2]2
= a4r4(1 + r2)2
and (a2 + b2) (c2 + d2)
= (a2 + a2r2) (a2r4 + a2r6)
= a2(1 + r2) a2r4(1 + r2)
= a4r4(1 + r2)2
∴ (b2 + c2)2 = (a2 + b2) (c2 + d2)
Thus a2 + b2, b2 + c2 and c2 + d2 are in G.P.
Que-7: If a, b, c, d are in G.P., prove that
(i) (b + c)(b + d) = (c + a)(c + d)
(ii) (a – d)2 = (b – c)2 + (c – a)2 + (d – b)2
Sol: Given a, b, c and d are in G.P
∴ b/a = c/b = d/c = r ≠ 0 (say)
∴ b = ar; c = br = ar2; d = cr = ar3
(i) L.H.S = (b + c)(b + d)
= (ar + ar2) (ar + ar3)
= ar(1 + r) ar(1 + r2)
= a2r2(1 + r) (1 + r2)
R.H.S = (c + a)(c + d)
= (ar2 + a) (ar2 + ar3)
= a(r2 + 1) ar2(1 + r)
= a2r2(1 + r) (1 + r2)
∴ L.H.S = R.H.S
(ii) L.H.S = (a – d)2 = (a – ar3)2
= a2 (1 – r3)2
R.H.S = (b – c)2 + (c – a)2 + (d – b)2
= (ar – ar2)2 + (ar2 – a)2 + (ar3 – ar)2
= (ar)2(1 – r)2 + a2(r2 – 1)2 + (ar)2 (r2 – 1)2
= a2[r2(1 – r)2 + (r2 – 1)2 + r2(r2 – 1)2]
= a2[r2(1 – r)2 + (r2 – 1)2 (1 + r2)]
= a2(1 – r)2 [r2 + (r + 1)2(1 + r2)]
= a2(1 – r)2[r2 + (r2 + 2r + 1) (1 + r2)]
= a2(1 – r)2[r2 + (1 + r2)2 + 2r (1 + r2)]
= a2(1 – r)2 (r + 1 + r2)2
= a2[(1 – r) (1 + r + r2)]2
= a2(1 – r3)2
∴ L.H.S = R.H.S
Que-8: If the pth, rth and rth terms of an A.P. are in G.P., prove that the common ratio of the G.P. is {(q-r)/(p-q)}
Sol: Let a be the first term and d be the common difference of A.P respectively.
Since Tp, Tq and Tr are in G.P.
∴ Common ratio = Tq / Tp = Tr / Tq
= {a+(q−1)d}/{a+(p−1)d} = {a+(r−1)d}/{a+(q−1)d}
= {(q−1−r+1)d}/{(p−1−q+1)d} = (q−r)/(p−q)
Que-9: If {1/(x+y)}, {1/2y}, {1/(y+z)} are the three consecutive terms of an A.P. prove that x, y, z are the three consecutive terms of a G.P.
Sol: Since {1/(x+y)}, (1/2y) and {1/(y+z)} are three consecutive terms of A.P.
⇒ (1/2y) – {1/(x+y)} = {1/(y+z)} – (1/2y)
⇒ {x+y−2y}/{2y(x+y)} = {2y−y−z}/{2y(y+z)}
⇒ {x−y}/{2y(x+y)} = {y−z}/{2y(y+z)}
⇒ (x – y)(y + z) = (x + y)(y – z)
⇒ xy + xz – y2 – yz = xy – xz + y2 – yz
⇒ 2y2 = 2xz ⇒ y2 = xz
Thus x, y and z are in G.P.
–: End of Sequence and Series Class 11 OP Malhotra Exe-14G ISC Maths Ch-14 Solutions. :–
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