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GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

If in the expansion of (1 + y)²⁰, the coefficients of r^th and (r + 4)^th terms are equal, then r is equal to

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જવાબ : 9

The term without y in the expansion of (2y − 1/2y²)¹² is

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જવાબ : 7920

The middle term in the expansion of (2y²/3 + 3/2y²)¹⁰ is

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જવાબ : 252

If an the expansion of (1+y)¹⁵, the coefficients of (2r+3)^th and (r−1)^th terms are equal, then the value of r is

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જવાબ : 5

In the expansion of (y²−1/3y)⁹, the term without y is equal to

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જવાબ : 28/243

The coefficient of x⁻¹⁷ in the expansion of (x⁴−1/x³)¹⁵ is

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જવાબ : −1365

The number of irrational terms in the expansion of (4^1/5+7^1/10)⁴⁵ is

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જવાબ : 41

If A and B are the sums of odd and even terms respectively in the expansion of (y + a)ⁿ, then (y + a)²ⁿ − (y − a)²ⁿ is equal to

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જવાબ : 4 AB

If in the expansion of (x + y)ⁿ and (x + y)ⁿ + 3, the ratio of the coefficients of second and third terms, and third and fourth terms respectively are equal, then n is

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જવાબ : 5

The coefficient of 1/y in the expansion of (1+y)ⁿ (1+1/y)ⁿ is

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જવાબ : 2n ! / [(n-1) ! (n+1) !]

If T₂/T₃ in the expansion of (a+b)ⁿ and T3/T4 in the expansion of (a+b)ⁿ⁺³ are equal, then n =

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જવાબ : 5

The total number of terms in the expansion of (y+a)¹⁰⁰+ (y−a)¹⁰⁰ after simplification is

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જવાબ : 51

The coefficient of x⁴ in (x/2 – 3/x²)¹⁰ is

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જવાબ : 405/256

If the coefficient of y in (y²+λ/y)⁵ is 270, then λ=

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જવાબ : 3

If the sum of odd numbered terms and the sum of even numbered terms in the expansion of (x+a)ⁿ are A and B respectively, then the value of (x²−a²)ⁿ is

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જવાબ : A²− B²

In the expansion of ( ½ y^1/3 + y^−1/5)⁸, the term independent of y is

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જવાબ : T₆

If in the expansion of (1 + x)ⁿ, the coefficients of 5th, 6th and 7th terms are in A.P., then n is equal to

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જવાબ : 7, 14

In the expansion of (y − 1/3y²)⁹, the term independent of y is

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જવાબ : T₄

If in the expansion of (y⁴− 1/y³)¹⁵, x⁻¹⁷ occurs in r^th term, then

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જવાબ : r = 12

The general term of the expansion (p + q)ⁿ is

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જવાબ : T_r+1 = ⁿC_r × p^n-r × q^r

If a and b are the roots of the equation y² – y + 1 = 0 then the value of a²⁰⁰⁹ + b²⁰⁰⁹ is

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જવાબ : 1

The value of n in the expansion of (p + q)ⁿ if the first three terms of the expansion are 729, 7290 and 30375, respectively is

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જવાબ : 6

The coefficient of yⁿ in the expansion of (1 – 2y + 3y² – 4y³ + ……..)^-n is

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જવાબ : (2n)!/(n!)²

The greatest coefficient in the expansion of (1 + y)¹⁰ is

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જવાબ : 10!/(5!)²

If the third term in the binomial expansion of (1 + y)ⁿ is (-1/8)y² then the rational value of n is

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જવાબ : 1/2

If m is a positive integer, then (√7+1)^2m+1 + (√7−1)^2m+1 is

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જવાબ : an irrational number

The fourth term in the expansion (a – 2b)¹² is

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જવાબ : -1760 a⁹ × b³

(1.2)¹⁰⁰⁰⁰ is _____ 1000

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જવાબ : greater than

The coefficient of x in the expansion of (x² + a/x)⁵ is

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જવાબ : 10a³

The number of ordered triplets of positive integers which are solution of the equation a + b + c = 100 is

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જવાબ : 4851

The greatest coefficient in the expansion of (1 + y)¹⁰ is

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જવાબ : 10!/(5!)²

if n is a positive ineger then 3³ⁿn – 3n is divisible by

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જવાબ : 49

In the expansion of (p + q)ⁿ, if n is odd then the number of middle term is/are

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જવાબ : 2

In the expansion of (p + q)ⁿ, if n is even then the middle term is

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જવાબ : (n/2 + 1)^th term

If n is a positive integer, then (√3+1)²ⁿ + 1 − (√3−1)²ⁿ + 1 is

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જવાબ : not an integer

The coefficient of yⁿ in the expansion (1 + y + y² + …..)^-n is

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જવાબ : (-1)ⁿ

If the coefficients of y² and y³ in the expansion of (3 + ay)⁹ are the same, then the value of a is

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જવાબ : 97

If the sum of the binomial coefficients of the expansion (2y + 1/y)ⁿ is equal to 256, then the term independent of y is

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જવાબ : 1120

If the fifth term of the expansion ( p^2/3 + p⁻¹)ⁿ does not contain 'p'. Then n is equal to

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જવાબ : 10

The coefficient of y⁻³ in the expansion of (y − n/y)¹¹ is

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જવાબ : −330 n⁷

The coefficient of the term independent of x in the expansion of (ax+bx)14 is

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જવાબ : 14!/(7!)² a⁷ b⁷

The coefficient of y⁵ in the expansion of (1+y)²¹+ (1+y)²²+ ... + (1+y)³⁰

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જવાબ : ³¹C₆ − ²¹C₆

The coefficient of p⁸ q¹⁰ in the expansion of (p + q)¹⁸ is

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જવાબ : ¹⁸C₈

If the coefficients of the (m + 1)^th term and the (m + 3)^th term in the expansion of (1 + y)²⁰ are equal, then the value of m is

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જવાબ : 9

If the coefficients of 2^nd, 3^rd and 4^th terms in the expansion of (1+y)ⁿ, n ∈ N are in A.P., then n =

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જવાબ : 7

The middle term in the expansion of (2y/3 – 3/2y²)²ⁿ is

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જવાબ : (−1)ⁿ ²ⁿC_n y⁻ⁿ

If r^th term is the middle term in the expansion of (y²−1/2y)²⁰, then (r+3)^th term is

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જવાબ : −²⁰C₇ y, 2⁻¹³

The number of terms with integral coefficients in the expansion of (17^1/3+ 35^1/2 y)⁶⁰⁰ is

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જવાબ : 101

Constant term in the expansion of (y−1/y)¹⁰ is

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જવાબ : -252

If r^th term in the expansion of (2y²−1/y)¹² is without y, then r is equal to

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જવાબ : 9

Find the middle terms(s) in the expansion of : (x/a – a/x)¹⁰

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જવાબ : (x/a + a/x)¹⁰

Here, n is an even number.

∴ Middle term = (10/2 +1 )^th= 6th term

Now, we have T₆ = T₅₊₁ = ¹⁰C₅ (x/a)¹⁰⁻⁵ (a/x)⁵ = 10×9×8×7×6/5×4×3×2×1 = 252

Find the middle terms(s) in the expansion of : (p/x + x/p)⁹

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જવાબ : (p/x - x/p)⁹

Here, n is an odd number.

Therefore, the middle terms are (9+1/2)^th and [(9+1)/2 + 1]^th, i.e., 5th and 6th terms.

Now, we have T₅ = T₄₊₁ = ⁹C₄ (p/x)⁹⁻⁴ (-x/p)⁴ = (9×8×7×6)/(4×3×2×1) × (p/x) = 126 p/x And,T₆ = T₅₊₁ = ⁹C₅ (p/x)⁹⁻⁵ (-x/p)⁵ = (9×8×7×6)/(4×3×2×1) × (-x/p) = -126 x/p

Find the middle terms(s) in the expansion of : (2ax−b/x²)¹²

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જવાબ : (2ax + b/x²)¹²

Here, n is an even number.

∴ Middle term = (12/2 +1)^th= 7th term

Now, we have T₇= T₆ + 1 = ¹²C₆ (2ax)¹²⁻⁶ (b/x²)⁶ = (12×11×10×9×8×7)/(6×5×4×3×2×1) × (2ab/x)⁶= 59136 a⁶b⁶/x⁶

Find the middle terms(s) in the expansion of : (3 − x³/6)⁷

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જવાબ : (3 + x³/6)⁷

Here, n is an odd number.

Therefore, the middle terms are (7+1/2)^th and [(7+1)/2 + 1]^th, i.e., 4th and 5th terms.

Now, we have T₄ = T₃₊₁ = ⁷C₃ (3)⁷⁻³ (x³/6)³

= 105/8 x⁹

And, T₅ = T₄₊₁ = ⁹C₄ (3)⁹⁻⁴ (x³/6)⁴ = (7×6×5)/(3×2) × 3⁵ × 1/6⁴ x¹²= 35/48 x¹²

Find the middle terms(s) in the expansion of : (x/3 + 9y)¹⁰

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જવાબ : (x/3 - 9y)¹⁰

Here, n is an even number.

Therefore, the middle term is (10/2 + 1 )^th, i.e., 6th term.

Now, we have T₆ = T₅₊₁ = ¹⁰C₅ (x/3)¹⁰⁻⁵ (-9y)⁵

= -(10×9×8×7×6)/(5×4×3×2) × 1/3⁵ × 9⁵ × x⁵ y⁵

= -61236 x⁵ y⁵

Find the middle terms(s) in the expansion of : (x – 1/x)²ⁿ⁺¹

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જવાબ : (x + 1/x)²ⁿ⁺¹

Here, (2n+1) is an odd number.

Therefore, the middle terms are (2n+1+1/2)^th and [(2n+1+1)/2 + 1]^th i.e. (n+1)^th and (n+2)^th terms.

Now, we have: T_n+1 = ²ⁿ⁺¹C_n x²ⁿ⁺¹⁻ⁿ × (1)ⁿ/xⁿ = (1)ⁿ ²ⁿ⁺¹C_n x

And,T_n+2 = T_n+1+1 = ²ⁿ⁺¹C_n+1 x^2n+1−n−1 (1)ⁿ⁺¹/xⁿ⁺¹ = (1)ⁿ⁺¹ ²ⁿ⁺¹C_n+1 × 1/x

Find the middle terms(s) in the expansion of : (2x − x²/4)⁹

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જવાબ : (2x + x²/4)⁹

Here, n is an odd number.

Therefore, the middle terms are [(n+1)/2]^th and [(n+1)/2 + 1]^th, i.e. 5th and 6th terms.

Now, we haveT₅ = T₄₊₁ = ⁹C₄ (2x)⁹⁻⁴ (x²/4)⁴ = (9×8×7×6)/(4×3×2) × 2⁵ 1/4⁴ x⁵⁺⁸ =63/4 x¹³

And,T₆ = T₅₊₁ = ⁹C₅ (2x)⁹⁻⁵ (x²/4)⁵ = (9×8×7×6)/(4×3×2) × 2⁴ 1/4⁵ x⁴ + 10 = 63/32 x¹⁴

Find the middle terms(s) in the expansion of : (1 + 3x + 3x²+ x³)²ⁿ

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જવાબ : (1 + 3x - 3x² + x³)²ⁿ =(1 - x)⁶ⁿ

Here, n is an even number.

∴ Middle term = (6n/2 + 1 )^th = (3n+1)^th term

Now, we have T_3n+1 = ⁶ⁿC_3n x³ⁿ = (6n)!/(3n!)²x³ⁿ

Find the middle terms(s) in the expansion of : (1 + 2x + x²)ⁿ

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જવાબ : (1 + 2x + x²)ⁿ

= (1+x)²ⁿ

n is an even number.

∴ Middle term = (2n/2 + 1)^th = (n+1)^th term

Now, we haveT_n+1 = ²ⁿC_n (1)ⁿ (x)ⁿ = (2n)!/(n!)² (1)ⁿ xⁿ

Find the middle terms(s) in the expansion of : (x + 1/x)¹⁰

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જવાબ : (x + 1/x)¹⁰

Here, n is an even number.

∴ Middle term = (10/2 + 1)^th= 6th term

Now, we haveT₆= T₅+ 1 =¹⁰C₅ x¹⁰⁻⁵ (1/x)⁵ = (10×9×8×7×6)/(5×4×3×2) = 252

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