MATHEMATICS XI TEST PAPER SOL.pdf

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Find the number of terms in the expansion of (a + 2b – 3c)n.
Consider (a + 2b – 3c)n = {a + (2b – 3c)}n = nC0 an + nC1 an–1(2b – 3c) + nC2 an–2(2b –
3c)2 + ... + nCn(2b – 3c)n.
We notice first term gives one term, second term gives two terms, third term gives three
terms and so on.
∴ Total terms 1 + 2 + ... + n + (n + 1) =
Find the middle term(s) in the expansion of : (1 + x)2n
Since, 2n would always be even so (n + 1)th term would be the middle term
∴
The coefficients of 2nd, 3rd and 4th terms in the expansion of (1 + x)2n are in A.P., prove that 2n2
– 9n + 7 = 0
2nC1, 2nC2, 2nC3 are in A.P.
⇒ 2. 2nC2 = 2nC1 + 2nC3
⇒
⇒ 2n2 – 9n + 7 = 0
If Tr is rth term in the expansion of (1 + x)n in the ascending powers of x, prove that r (r + 1) Tr+2
= (n – r + 1) (n – r)x2 Tr.
LHS = r (r + 1) Tr+2 = r (r + 1). nCr+1 xr+1
= r (r + 1)
=
RHS = (n – r + 1) (n – r) x2 Tr
= (n – r + 1) (n – r)x2. nCr–1 xr–1
= (n – r + 1) (n – r). x2 .
=
LHS = RHS

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If the 3rd, 4th, 5th and 6th terms in the expansion of (x + y)n be a, b, c and d respectively, prove
that
If a, b, c be the three consecutive coefficients in the expansion of (1 + x)n, prove that n =

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7.
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8.
Using binomial theorem, prove that 6n – 5n always leaves remainder 1 when divided by 25.
For two numbers a and b if we can find numbers q and r such that a = bq + r, then we
say that b divides a with q as quotient and r as remainder. Thus, in order to show that 6n
– 5n leaves remainder 1 when divided by 25, we prove that 6n – 5n = 25k + 1, where k is
some natural number.
We have
(1 + a)n = nC0 + nC1a + nC2a2 + ... nCnan
For a = 5, we get
(1 + 5)n = nC0 + nC15 + nC252 + ... nCn5n
i.e. (6)n = 1 + 5n + 52 nC2 + 53 nC3 + ... + 5n
i.e. 6n – 5n = 1 + 52 (nC2 + nC3 5 + ... + 5n–2)
or 6n – 5n = 1 + 25 (nC2 + 5. nC3 + ... + 5n–2)
or 6n – 5n = 25k + 1
where k = nC2 + 5. nC3 + ... + 5n–2
This shows that when divided by 25, 6n – 5n leaves remainder 1.
If a1, a2, a3 and a4 are the coefficient of any four consecutive terms in the expansion of (1 + x)n,
prove that

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9.
Let a1, a2, a3 and a4 be the coefficient of four consecutive terms Tr+1, Tr+2, Tr+3 and Tr+4
respectively. Then
a1 = coefficient of Tr+1 = nCr
a2 = coefficient of Tr+2 = nCr+1
a3 = coefficient of Tr+3 = nCr+2
and a4 = coefficient of Tr+4 = nCr+3
Find the sum (33 – 23) + (53 – 43) + (73 – 63) + ... 10 terms.

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Sum = (33 + 53 + 73 + ...) – (23 + 43 + 63 + ...)
For 33 + 53 + 73 + ...
an = (2n + 1)3 = 8n3 + 12n2 + 6n + 1
∴ S1 = 8 Σn3 + 12 Σn2 + 6 Σn + n ... (i)
For 23 + 43 + 63 + ...
an = (2n)3 = 8n3
S2 = 8Σn3 ... (ii)
∴ Sum = S1 – S2 = 12 Σn2 + 6 Σn + n [From (i), (ii)]
Sum to 10 terms =
= 4620 + 330 + 10 = 4960
If a, b, c be the first, third and nth term respectively of an A.P. Prove that the sum to n terms is
Given first term = a, a3 = b and an = c
Sn = [2a + (n – 1)d] ... (i)
We have first term = a
b = a + 2d and a + (n – 1) d = c
Sn = [a + c] [From (i)] ... (ii)
Also a + (n – 1) = c
⇒ n – 1 =
Substituting in (ii), we get
Sn =
If a, b, c are in G.P and . Prove that x, y, z are in A.P.
⇒ a = kx, b = ky, c = kz
Given b2 = ac ⇒ (ky)2 = kx. kz ⇒ k2y = kx + z
⇒ 2y = x + z ⇒ x, y, z in A.P.
If a1, a2, a3, ... an are in A.P., where ai > 0 for all i, show that
=

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13. The ratio of the A.M. and G.M. of two positive numbers a and b is m : n, show that:

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14.
Given A.M. : G.M. = m : n
If p, q, r are in G.P. and the equations px2 + 2qx + r = 0 and dx2 + 2ex + f = 0 have a common
root then show that are in A.P.

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Given p, q, r are in G.P.
⇒ q2 = pr …(i)
Equations px2 + 2qx + r = 0 and dx2 + 2ex + f = 0
have common root …(ii)
Substituting for q, from (i) in px2 + 2qx + r = 0
Find the sum of, + .......................... up to n terms.
If S1, S2, S3, ......, Sm are the sums of n terms of m A.P.'s whose first terms are 1, 2, 3, ......, m
and common differences are 1, 3, 5, ......, 2m – 1 respectively, show that S1 + S2 + S3 + ...... + Sm =

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17.
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mn (mn + 1)
S1 = Σ(1, 2, 3..............) = [2.1 + (n – 1).1]
S2 = Σ(2, 5, 8..............) = [2.2 + (n – 1).3]
S3 = Σ(3, 8, 13............) = [2.3 + (n – 1).5]
Sm = [2.m + (n – 1) (2m – 1)]
S1 + S2 + ...... + Sm = [2 (1 + 2 + 3.......m) + (n – 1) (1 + 3 + 5........+ 2m – 1)]
=
=
If A and G be A.M. and G.M., respectively between two positive numbers, prove that the
numbers are A ± .
Let the numbers be a and b.
= A ⇒ a + b = 2A ...(i)
= G ⇒ ab = G2
(a – b)2 = (a + b)2 – 4ab = 4A2 – 4G2
⇒ 4(A2 – G2)
a – b = ...(ii)
From (i) and (ii), we get
a = A + and b = A –
Hence, the numbers are A ±

MATHEMATICS XI TEST PAPER SOL.pdf

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MATHEMATICS XI TEST PAPER SOL.pdf