Sequence and Series Class 11 OP Malhotra Exe-14H ISC Maths Solutions Ch-14 Solutions. In this article you would learn about Arithmetico-Geometric Series. 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-14H 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(H)	Arithmetico-Geometric Series

Arithmetic Geometric Series

Sequence and Series Class 11 OP Malhotra Exe-14H ISC Maths Solutions Ch-14 Solutions

Que-1: 1 + 2x + 3x² + 4x³ + ………

Sol: Let S_n = 1 + 2x + 3x² + 4x³ + ……… n terms
i.e. S_n = 1 + 2x + 3x² + 4x³ + …..,.+(n-1) x^{n-2}+n x^{n-1}$
It is an arithmetic-geometric series.
x S_n = x + 2x² + 3x³ + ………. + (n + 1) x^{n – 1} + nxⁿ
eqn. (1) – eqn. (2) gives;
(1 – x)S_n = 1 + x + x² + ……. + x^{n – 1} – nxⁿ = [{1(1−x^n)}/{1−x}] – nx^n
⇒ Sn = {1−x^n}/{(1−x)² – (nx^n)/(1−x)

Que-2: 1 + 3x +5x² + 7x³ + …..

Sol: Let S_n = 1 + 3x + 5x² + 7x³ + …… +(2n – 3) x^n-2 + (2n – 1) x^n-1 …(1)
It is an A.G.P.
∴ xS_n = x + 3x² + 5x³ + ….. +(2n – 3) x^n-1 + (2n – 1) xⁿ …(2)
eqn. (1) – eqn. (2) gives ;
(1 – x)S_n = 1 + 2x + 2x² + 2x³ – …. + 2x^n-1 – (2n – 1) xⁿ
= 1 + [2x[1−x^(n−1)]]/(1−x)  –  {(2n – 1)x^n}
= 1 + {2x/(1−x)} – {(2x^n)/(1−x)} – (2n – 1)x^n
∴ Sn = {(1+x)/(1−x)²} – {(2x^n)/(1−x)²} – {(2n−1)x^n}/(1−x)

Que-3: 2.1 + 3.2 + 4.4 + 5.8 + ….

Sol: Let S_n = 2.1 + 3.2 + 4.4 + 5.8 + …… n 2^n-2+(n+1) 2^n-1 …(1)
∴ 2S_n = 2.2 + 3.4 + 4.8 + …….. + n 2^n-1 + (n+1) 2ⁿ …(2)
eqn. (1) – eqn. (2) gives ;
-S_n = 2.1 + 1.2 + 1.4 + 1.8 ……. + 2^n-1 – (n + 1) 2ⁿ = 2 + {2(2^(n−1)−1)}/(2−1) – (n + 1)2ⁿ
⇒ -S_n = 2 + 2 (2^n-1 – 1) – (n + 1) 2ⁿ = 2 + 2ⁿ – 2 – n2ⁿ – 2ⁿ = -n2ⁿ
⇒ S_n = n.2ⁿ

Que-4: 1/2 + 3/6 + 5/18 + ……

Sol: Let S = 1/2 + 3/6 + 5/18 + …… ∞ …(1)
It is an arithmetic geometric series.
(1/3)S = (1/6) + (1/18) + ……….. ∞ …(2)
eqn (1) – eqn (2) gives

S = 3/2

Que-5: 3/2 – 5/6 + 7/18 – ……

Sol: Let S = 3/2 – 5/6 + 7/18  – …… ∞ …(1)
It is an arithmetic geometric series with r = −1/3
∴ −(1/3)S = (−3/6) + (5/18) + ….. ∞ …(2)
eqn. (1) – eqn. (2) gives ;
(4/3)S = (3/2) – (2/6) + (2/18) + ….. ∞
= (3/2) + {(−2/6)/{1+(1/3)}
= (3/2) + (−1/3) × (3/4)
= (3/2) – (1/4) = 5/4
⇒ S = (5/4) × (3/4) = 15/16

Que-6: 1 – 2/5 + 3/5² – 4/5³ + …..

Sol: Let S = 1 – 2/5 + 3/5² – 4/5³ + ….. ∞ …(1)
∴ – (1/5)S = – (1/5) + (2/5²) – (3/5³) + ….. ∞ …(2)
eqn. (1) – eqn. (2) gives ;
(6/5)S = 1 – (1/5) + (1/5²) …… ∞
= [1/{1+(1/5)}]
⇒ (6/5)S = (5/6)
⇒ S = 25/36

Que-7: Sum up the series 2/3 + 5/9 + 8/27 + 11/81 + …. to n terms and hence find the sum to infinity.

Sol: Let 2/3 + 5/9 + 8/27 + 11/81 + ….  n terms

from (3); we have,
S∞ = 7/4

Que-8: 1 + 4x² + 7x⁴ + ……

Sol: Let S = 1 + 4x² + 7x⁴ + ……… ∞ …(1)
x²S = x² + 4x⁴ + ……… ∞ …(2)
eqn. (1) – eqn. (2) gives ;
(1 – x)² S = 1 + 3x² + 3x⁴ + …….. ∞
= 1 + {3x²/(1−x²)} = (1−x²+3x²)/(1−x²)
= (1+2x²)/(1−x²)
⇒ S = (1+2x²)/(1−x²)²

Que-9: 1 – x + 2x² – 3x³ + 4x⁴ – …………

Sol: Let S = 1 – x + 2x² – 3x³ + …… ∞
⇒ S = 1 – {x – 2x² + 3x³ + …… ∞} …(1)
which is an A.G.P with common ratio (-x).
∴ -xS = -x + x² – 2x³ + …… ∞ …(2)
eqn. (1) – eqn. (2) gives ;
(1 + x)S = 1 + x² – x³ – …… ∞
= 1 + {x²/(1+x)}
⇒ S = {1/(1+x)} + {x²/(1+x)²}
= (1+x+x²)/(1+x)²

Que-10: 1² + 3²x + 5²x² + 7²x³ + …..

Sol: Let S = 1² + 3²x + 5²x² + 7²x³ + …… ∞
∴ S = 1 + 9x + 25x² + 49x³ + …… ∞ …(1)
xS = x + 9x² + 25x³ + …… ∞ …(2)
eqn. (1) – eqn. (2) gives ;
(1 – x) S = 1 + 8x + 16x² + 24x³ + …… ∞ …(3)
It is clearly an A.G.P series.
∴ x(1 – x)S = x + 8x² + 16x³ + …… ∞ …(4)
eqn. (3) – eqn. (4) gives;
(1 – x)²S = 1 + 7x + 8x² + 8x³ + …… ∞
= 1 + 7x + {8x²/(1−x)}
⇒ S = {(1+7x)/(1−x)²} + {8x²/(1−x)³}

Que-11: Show that the square root of {3^(1/2)} × {9^(1/4)} × {27^(1/8)} × {81^(1/16)} × ……. to infinity is 3 .

Sol: Let S = {3^(1/2)} × {9^(1/4)} × {27^(1/8)} × {81^(1/16)} × ……. ∞

from (1) ; S = 3² = 9
Thus square root of S = √9 = 3.

–: End of Sequence and Series Class 11 OP Malhotra Exe-14H ISC Maths Ch-14 Solutions. :–

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