Free Objective Test 03 Practice Test - 11th and 12th

Question 1

If $π \sum i = 1 i$ = $\frac{n (n + 1)}{2}$ , then $π \sum i = 1 (3 i - 2)$ =

$\frac{n (3 n - 1)}{2}$

$\frac{n (3 n + 1)}{2}$

n(3n + 2)

$\frac{n (3 n + 1)}{4}$

SOLUTION

Solution : A

$π \sum i = 1$ = 3 $π \sum i = 1 i$ - 2 $π \sum i = 1 1$ = 3 $\frac{n (n + 1)}{2}$ - 2n = $\frac{n (3 n - 1)}{2}$

Question 2

$\frac{1}{1.2}$ + $\frac{1}{2.3}$ + $\frac{1}{3.4}$ +.......+ ......... $\frac{1}{n . (n + 1)}$ equals

$\frac{1}{n (n + 1)}$

$\frac{n}{n + 1}$

$\frac{2 n}{n + 1}$

$\frac{2}{n (n + 1)}$

SOLUTION

Solution : B

( $\frac{1}{1}$ - $\frac{1}{2}$ ) + ( $\frac{1}{2}$ - $\frac{1}{3}$ ) + ( $\frac{1}{3}$ - $\frac{1}{4}$ ) + .........+ ( $\frac{1}{n}$ - $\frac{1}{n + 1}$ )

= 1 - $\frac{1}{n + 1}$ = $\frac{n}{n + 1}$

Question 3

The sum of n terms of the series whose $n^{t h}$ term is n(n+1) is equal to

$\frac{n (n + 1) (n + 2)}{3}$

$\frac{(n + 1) (n + 2) (n + 3)}{12}$

$n^{2}$ (n+2)

n(n+1)(n+2)

SOLUTION

Solution : A

$T_{π}$ = $n^{2}$ + n ⇒ $S_{π}$ = $\sum T_{π}$ = $\sum$ $n^{2}$ + $\sum$ n

= $\frac{n (n + 1) (2 n + 1)}{6}$ + $\frac{n (n + 1)}{2}$

= $\frac{n (n + 1)}{6}$ {2n + 1 + 3} = $\frac{n (n + 1) (n + 2)}{3}$

Question 4

If $a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}}$ and $a, b, c$ are in G.P., then which of the following are true?

$x, y, z$ will be in A.P.

$x, y, z$ will be in G.P.

$x, y, z$ will be in H.P.

D. None of the above

SOLUTION

Solution : A

Let $a^{\frac{1}{x}} = b^{\frac{1}{y}} = c^{\frac{1}{z}} = k .$
$\Rightarrow a = k^{x}, b = k^{y}, c = k^{z} .$

Since $a, b, c$ are in G.P., $b^{2} = a c .$
$∴ k^{2 y} = k^{x} . k^{z} = k^{x + z}$
$∴ 2 y = x + z$
$\Rightarrow x, y, z$ are in A.P.

Question 5

If $\frac{1}{b - c}$ , $\frac{1}{c - a}$ , $\frac{1}{a - b}$ be consecutive terms of an A.P., then $(b - c)^{2}, (c - a)^{2}, (a - b)^{2}$ will be in ___.

G.P

A.P

H.P

None of these

SOLUTION

Solution : B

If $(b - c)^{2}, (c - a)^{2}, (a - b)^{2}$ are in A.P.

Then we have $(c - a)^{2} - (b - c)^{2}$ = $(a - b)^{2} - (c - a)^{2}$

$\Rightarrow (b - a) (2 c - a - b)$ = (c - b)(2a - b - c) ......(i)

Also if $\frac{1}{b - c}, \frac{1}{c - a}, \frac{1}{a - b}$ are in A.P.

Then $\frac{1}{c - a} - \frac{1}{b - c}$ = $\frac{1}{a - b} - \frac{1}{c - a}$

$\Rightarrow \frac{b + a - 2 c}{(c - a) (b - c)}$ = $\frac{c + b - 2 a}{(a - b) (c - a)}$

$\Rightarrow (a - b) (b + a - 2 c)$ = $(b - c) (c + b - 2 a)$

$\Rightarrow (b - a) (2 c - a - b)$ = $(c - b) (2 a - b - c)$

Which is true by virtue of (i).

Question 6

If $p^{t h}$ , $q^{t h}$ , $r^{t h}$ and $s^{t h}$ terms of an A.P. be in G.P., then (p - q),(q - r),(r - s) will be in

G.P

A.P

H.P

None of these

SOLUTION

Solution : A

If a and d be the first term and common difference of the A.P.

Then $T_{p}$ = a + (p - 1)d, $T_{q}$ = a + (q - 1)d and

$T_{r}$ = a+(r - 1)d.

If $T_{p}, T_{q}, T_{r}$ are in G.P.

Then its common ratio R = $\frac{T_{q}}{T_{p}}$ = $\frac{T_{r}}{T_{q}}$ = $\frac{T_{q} - T_{r}}{T_{p} - T_{q}}$

= $\frac{[a + (q - 1) d] - [a + (r - 1) d]}{[a + (p - 1) d] - [a + (q - 1) d]}$ = $\frac{q - r}{p - q}$

Similarly, we can show that R = $\frac{q - r}{p - q}$ = $\frac{r - s}{q - r}$

Hence (p - q), (q - r),(r - s) be in G.P.

Question 7

If the numbers be in G.P., then their logarithms will be in

A.P

G.P

H.P

None of these

SOLUTION

Solution : A

Let three number a,b and c in G.P., then $b^{2}$ = ac

$\Rightarrow 2 l o g_{d} b$ = $l o g_{d} a + l o g_{d} c$ or $l o g_{d} b$ = $\frac{l o g_{d} a + l o g_{d} c}{2}$

Thus their logarithms are in A.P.

Question 8

If $l o g_{a} x, l o g_{b} x, l o g_{c} x$ be in H.P., then a,b,c are in

A.P

H.P

G.P

None of these

SOLUTION

Solution : C

Here $\frac{l o g x}{l o g a}, \frac{l o g x}{l o g b}, \frac{l o g x}{l o g c}$ are in H.P.

$\Rightarrow \frac{l o g a}{l o g x}, \frac{l o g b}{l o g x}, \frac{l o g c}{l o g x}$ are in A.P.

$\Rightarrow l o g_{x} a, l o g_{x} b, l o g_{x} c$ are in A.P.

$\Rightarrow a, b, c$ are in G.P.

Note: Students should remember this question as afact

Question 9

The fifth of the H.P. 2, $\frac{5}{2}$ , $\frac{10}{3}$ ,................. will be

$5 \frac{1}{5}$

$3 \frac{1}{5}$

$\frac{1}{10}$

SOLUTION

Solution : D

Series, 2, $\frac{5}{2}$ , $\frac{10}{3}$ ,................. are in H.P. $\Rightarrow \frac{1}{2}, \frac{2}{5}, \frac{3}{10}$ ,......... will be in A.P.

Now first term a = $\frac{1}{2}$ and common difference d = $- \frac{1}{10}$

So, $5^{t h}$ term of the A.P. = $\frac{1}{2} + (5 - 1) (- \frac{1}{10})$ = $\frac{1}{10}$

Hence $5^{t h}$ term in H.P. is 10.

Question 10

If the $m^{t h}$ term of a H.P. be n and $n^{t h}$ be m, then the $r^{t h}$ term will be

$\frac{r}{m n}$

$\frac{m n}{r + 1}$

$\frac{m n}{r}$

$\frac{m n}{r - 1}$

SOLUTION

Solution : C

Given $T_{m}$ = n, $T_{n}$ = m for H.P. therefore for the corresponding A.P. $m^{t h}$ term = $\frac{1}{n}$ , $n^{t h}$ term = $\frac{1}{m}$

Let a and d be the first term and common difference of this A.P., then

$a + (m - 1) d$ = $\frac{1}{n}$ ......(i)

$a + (n - 1) d$ = $\frac{1}{m}$ ......(i)

Solving these, we get a = $\frac{1}{m n}$ , d = $\frac{1}{m n}$

Now, $r^{t h}$ term of corresponding A.P.

= $a + (r - 1) d$ = $\frac{1}{m n} + (r - 1) \frac{1}{m n}$ = $\frac{1 + r - 1}{m n}$ = $\frac{r}{m n}$

Therefore $r^{t h}$ term of corresponding H.P. is $\frac{m n}{r}$

Note : Students should remember this question as a fact.

Question 11

If a, b, c are $p^{t h}$ , $q^{t h}$ , and $r^{t h}$ terms of a G.P., then $(\frac{c}{b})^{p} (\frac{b}{a})^{r} (\frac{a}{c})^{q}$ is equal to

$a^{p} b^{q} c^{r}$

$a^{q} b^{r} c^{p}$

$a^{r} b^{p} c^{q}$

SOLUTION

Solution : A

a = $A R^{p - 1}$ , b = $A R^{q - 1}$ , c = $A R^{r - 1}$

$∴ (\frac{c}{b})^{p} (\frac{b}{a})^{r} (\frac{a}{c})^{q}$ = $(\frac{A R^{p - 1}}{A R^{q - 1}})^{p} (\frac{A R^{q - 1}}{A R^{p - 1}})^{r} (\frac{A R^{p - 1}}{A R^{r - 1}})^{q}$

= $R^{(r - q) p + (q - p) r + (p - r) q}$ = $R^{0}$ = 1

Question 12

If the $5^{t h}$ term of a G.P. containing positive terms is $\frac{1}{3}$ and $9^{t h}$ term is $\frac{16}{243}$ , then the $4^{t h}$ term will be

$\frac{3}{4}$

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{2}{5}$

SOLUTION

Solution : B

$T_{5}$ = $a r^{4}$ = $\frac{1}{3}$ .......(i)

and $T_{9}$ = $a r^{8}$ = $\frac{16}{243}$ .......(ii)

Solving (i) and (ii), we get r = $\frac{2}{3}$ and a = $\frac{27}{16}$

Now $4^{t h}$ term = $a r^{3}$ = $\frac{3^{3}}{2^{4}} . \frac{2^{3}}{3^{3}}$ = $\frac{1}{2}$

Question 13

The number which should be added to the number added to the numbers 2, 14, 62 so that the resulting numbers may be in G.P. is

SOLUTION

Solution : B

Suppose that the added number be x

then x + 2, x + 14, x + 62 be in G.P.

Therefore $(x + 14)^{2}$ = (x + 2)(x + 62)

$\Rightarrow x^{2} + 196 + 28 x$ = $x^{2} + 64 x + 124$

$\Rightarrow 36 x$ = 72 $\Rightarrow x$ = 2

Trick : (a) Let 1 is added, then the numbers will be 3, 15, 63 which are obviously not in G.P.

(b) Let 2 is added, then the numbers will be 4, 16, 64 which are obviously in G.P.

Question 14

If a, b, c, d are in H.P., then ab + bc + cd is equal to

3ad

(a + b)(c + d)

3ac

None of these

SOLUTION

Solution : A

Since a, b, c, d are in H.P., therefore b is the H.M. of a and c

i.e., b = $\frac{2 a c}{b + d}$ , $∴ (a + c) (b + d)$ = $\frac{2 a c}{b} . \frac{2 b d}{c}$

$\Rightarrow a b + a d + b c + c d$ = 4ad $\Rightarrow a b + b c + c d$ = 3ad

Trick : Check for a =1, b = $\frac{1}{2}$ , c = $\frac{1}{3}$ , d = $\frac{1}{4}$

Question 15

If the roots of $a (b - c) x^{2} + b (c - a) x + c (a - b)$ = 0 be equal then a,b,c are in

A.P

G.P

H.P

None of these

SOLUTION

Solution : C

Since the roots of the quadratic equation in x are equal, we have $(B^{2} - 4 A C)$ = 0

$\Rightarrow b^{2} (c - a)^{2} - 4 a c (b - c) (a - b)$ = 0

$\Rightarrow b^{2} (c^{2} - 2 c a + a^{2}) - 4 a c (b a - b^{2} - c a + b c)$ = 0

$\Rightarrow b^{2} (c^{2} + 2 c a + a^{2}) - 4 a c {b (a + c) - a c}$ = 0

$\Rightarrow b^{2} (a + c)^{2} - 4 a c {b (a + c) - a c}$ = 0

Which can be seen to be true, if

b = $\frac{2 a c}{a + c}$ or b(a+c) = 2ac i.e., if a,b,c are in H.P.

Free Objective Test 03 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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