Select CBSE Solutions for class 10 Subject & Chapters Wise :
If in the expansion of (1 + y)20, the coefficients of rth and (r + 4)th terms are equal, then r is equal to
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9
The term without y in the expansion of (2y − 1/2y2)12 is
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7920
The middle term in the expansion of (2y2/3 + 3/2y2)10 is
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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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5
In the expansion of (y2−1/3y)9, the term without y is equal to
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28/243
The coefficient of x−17 in the expansion of (x4−1/x3)15 is
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−1365
The number of irrational terms in the expansion of (41/5+71/10)45 is
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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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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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5
The coefficient of 1/y in the expansion of (1+y)n (1+1/y)n is
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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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5
The total number of terms in the expansion of (y+a)100 + (y−a)100 after simplification is
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51
The coefficient of x4 in (x/2 – 3/x2)10 is
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405/256
If the coefficient of y in (y2+λ/y)5 is 270, then λ=
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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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A2 − B2
In the expansion of ( ½ y1/3 + y−1/5)8, the term independent of y is
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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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7, 14
In the expansion of (y − 1/3y2)9, the term independent of y is
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T4
If in the expansion of (y4 − 1/y3)15, x−17 occurs in rth term, then
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r = 12
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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1120
If the fifth term of the expansion ( p2/3 + p−1)n does not contain 'p'. Then n is equal to
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10
The coefficient of y−3 in the expansion of (y − n/y)11 is
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−330 n7
The coefficient of the term independent of x in the expansion of (ax+bx)14 is
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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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31C6 − 21C6
If rth term in the expansion of (2y2−1/y)12 is without y, then r is equal to
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9
Find the middle terms(s) in the expansion of : (x – 1/x)2n+1
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(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: Tn+1 = 2n+1Cn x2n+1−n × (1)n/xn = (1)n 2n+1Cn x
And,Tn+2 = Tn+1+1 = 2n+1Cn+1 x2n+1−n−1 (1)n+1/xn+1 = (1)n+1 2n+1Cn+1 × 1/x
Find the middle terms(s) in the expansion of : (2x − x2/4)9
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(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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(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 T3n+1 = 6nC3n x3n = (6n)!/(3n!)2 x3n
Find the middle terms(s) in the expansion of : (1 + 2x + x2)n
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(1 + 2x + x2)n
= (1+x)2n
n is an even number.
∴ Middle term = (2n/2 + 1)th = (n+1)th term
Now, we haveTn+1 = 2nCn (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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(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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The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.
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