Home Previous CPT Papers(1993-2018) Quantitative Techniques Exam paper Sequence and Series – Test 2

Sequence and Series – Test 2

sequence and series

Subject :- QT

Chapter :- Sequence and Series-Test 2

Questions :- 25

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Question 1 of 25

1. Question
1 points
Find the sum of infinity of the following series
$1 - 1 + 1 - 1 + 1 - 1 + . . . . . . . . . . . . . . . . . . . . . . . . . . \infty$
- 1
- 0
- 1/2
- none of these
Correct

a=1

r=-1

sum of GP upto infinity=a/1-r=1/1-(-1)=1/1+1=1/2

Incorrect

a=1

r=-1

sum of GP upto infinity=a/1-r=1/1-(-1)=1/1+1=1/2
Question 2 of 25

2. Question
1 points
Divide 144 into 3 parts which are in AP and such the largest is twice the smallest ,the smallest of 3 numbers will be:
- 43
- 36
- 13
- 32
Correct

let the three numbers in AP be a-d,a,a+d

given,a+d=2(a-d)

a-d+a+a+d=144

3a=144

a=48

48+d=2(48-d)

48+d=96-2d

3d=48

d=16

smallest number=a-d

=48-16

=32

Incorrect

let the three numbers in AP be a-d,a,a+d

given,a+d=2(a-d)

a-d+a+a+d=144

3a=144

a=48

48+d=2(48-d)

48+d=96-2d

3d=48

d=16

smallest number=a-d

=48-16

=32
Question 3 of 25

3. Question
1 points
$S u m o f s e r i e s 1 + \frac{4}{5} + \frac{7}{5^{2}} + \frac{10}{5^{3}} + . . . . . . . . . . . . . . \infty i s$
- 15/36
- 35/36
- 35/16
- 15/16
Correct

Incorrect
Question 4 of 25

4. Question
1 points
$I f G b e G M b e t w e e n 2 n u m b e r s a a n d b t h e n t h e v a l u e o f \frac{1}{G^{2} - a^{2}} + \frac{1}{G^{2} - b^{2}} =$
- GxG
- 3/GxG
- 1/GxG
- 2/GxG
Correct

$\frac{1}{G^{2} - a^{2}} + \frac{1}{G^{2} - b^{2}} G M = \sqrt{a b} \frac{1}{a b - a^{2}} + \frac{1}{a b - b^{2}} = \frac{1}{a (b - a)} + \frac{1}{b (a - b)} = \frac{b - a}{a b (b - a)} = \frac{1}{a b} = \frac{1}{(\sqrt{a b})^{2}} = 1 / G^{2}$
=1/GxG

Incorrect

$\frac{1}{G^{2} - a^{2}} + \frac{1}{G^{2} - b^{2}} G M = \sqrt{a b} \frac{1}{a b - a^{2}} + \frac{1}{a b - b^{2}} = \frac{1}{a (b - a)} + \frac{1}{b (a - b)} = \frac{b - a}{a b (b - a)} = \frac{1}{a b} = \frac{1}{(\sqrt{a b})^{2}} = 1 / G^{2}$
=1/GxG
Question 5 of 25

5. Question
1 points
Find the product of
$243, 243^{1 / 6}, 243^{1 / 36}, . . . . . . . . . . . . . . . \infty$
- 1024
- 270
- 729
- 246
Correct

$p r o d u c t = 243, 243^{1 / 6}, 243^{1 / 36}, . . . . . . . . . . . . . . . \infty 1 + 1 / 6 + 1 / 36 + . . . . . . . \infty = 243 = 243^{\frac{1}{1 - 1 / 6}} = 243^{6 / 5} = (3^{5})^{6 / 5} = 3^{6} = 729$

Incorrect

$p r o d u c t = 243, 243^{1 / 6}, 243^{1 / 36}, . . . . . . . . . . . . . . . \infty 1 + 1 / 6 + 1 / 36 + . . . . . . . \infty = 243 = 243^{\frac{1}{1 - 1 / 6}} = 243^{6 / 5} = (3^{5})^{6 / 5} = 3^{6} = 729$
Question 6 of 25

6. Question
1 points
Insert 2 AM’s between 68 and 260
- 132,196
- 130,194
- 70,258
- none of these
Correct

$l e t A_{1}, A_{2} b e t h e 2 A m ‘ s b e t w e e n 68 a n d 260 68, A_{1}, A_{2}, 260 a r e i n A p h e r e, a = 68, l = 260 a n d n = 4 l = a + (n - 1) d 260 = 68 + (4 - 1) d d = 64 A_{1} = a + d = 68 + 64 = 132 A_{2} = a + 2 d = 68 + 2 x 64 = 196$

Incorrect

$l e t A_{1}, A_{2} b e t h e 2 A m ‘ s b e t w e e n 68 a n d 260 68, A_{1}, A_{2}, 260 a r e i n A p h e r e, a = 68, l = 260 a n d n = 4 l = a + (n - 1) d 260 = 68 + (4 - 1) d d = 64 A_{1} = a + d = 68 + 64 = 132 A_{2} = a + 2 d = 68 + 2 x 64 = 196$
Question 7 of 25

7. Question
1 points
Find the numbers whose AM is 12.5 and GM is 10
- 20,5
- 10,5
- 5,4
- none of these
Correct

$l e t t h e 2 n u m b e r s b e a a n d b A M = a + b / 2 = 12.5 a + b = 25 . . . . . . . (1) G m = \sqrt{a b} = 10 a b = 100 . . . . . . . . (2) (a - b)^{2} = (a + b)^{2} . 4 a b = 25^{2} - 4 x 100 = 625 - 400 = 225 = \sqrt{225} a - b = 15 . . . . . . . . (3) b y s o l v i n g 1 & 3 w e g e t a = 20, b = 5$

Incorrect

$l e t t h e 2 n u m b e r s b e a a n d b A M = a + b / 2 = 12.5 a + b = 25 . . . . . . . (1) G m = \sqrt{a b} = 10 a b = 100 . . . . . . . . (2) (a - b)^{2} = (a + b)^{2} . 4 a b = 25^{2} - 4 x 100 = 625 - 400 = 225 = \sqrt{225} a - b = 15 . . . . . . . . (3) b y s o l v i n g 1 & 3 w e g e t a = 20, b = 5$
Question 8 of 25

8. Question
1 points
If sum of 3 AM’s between a and 22 is 42,then a=____
- 14
- 11
- 10
- 6
Correct

Incorrect
Question 9 of 25

9. Question
1 points
If each month Rs 100 increases in any sum then find the total sum after 10 months,if the sum of the first month is Rs 2000
- Rs 24500
- Rs 24000
- Rs 50000
- Rs 60000
Correct

$a = 2000, d = 100 a n d n = 10 S_{n} = \frac{n}{2} [2 a + (n - 1) d] = 10 / 2 [2 x 2000 + (10 - 1) 100] = 5 [4000 + 900] = R s 24500$

Incorrect

$a = 2000, d = 100 a n d n = 10 S_{n} = \frac{n}{2} [2 a + (n - 1) d] = 10 / 2 [2 x 2000 + (10 - 1) 100] = 5 [4000 + 900] = R s 24500$
Question 10 of 25

10. Question
1 points
The sum of all 2 digit odd numbers is
- 2475
- 2575
- 4950
- 5049
Correct

Incorrect
Question 11 of 25

11. Question
1 points
The sum of the 3rd and 9th term of an AP is 8.find the sum of the first 11 terms of the progression.
- 44
- 22
- 19
- 11
Correct

Incorrect
Question 12 of 25

12. Question
1 points
If 8th term of an AP is 15,then sum of its 15 terms is
- 15
- 0
- 225
- 220
Correct

Incorrect
Question 13 of 25

13. Question
1 points
$I f t h e s u m o f n t e r m s o f a n A P b e 2 n^{2} + 5 n, t h e n i t s n t h t e r m i s$
- 4n-2
- 3n-4
- 4n+3
- 3n+4
Correct

$g i v e n, S_{n} = 2 n^{2} + 5 n S_{n - 1} = 2 (n - 1)^{2} + 5 (n - 1) = 2 n^{2} + 2 - 4 n + 5 n - 5 = 2 n^{2} + n - 3 n t h t e r m (T_{n}) = S_{n} - S_{n - 1} = (2 n^{2} + 5 n) - (2 n^{2} + n - 3) = 4 n + 3$

Incorrect

$g i v e n, S_{n} = 2 n^{2} + 5 n S_{n - 1} = 2 (n - 1)^{2} + 5 (n - 1) = 2 n^{2} + 2 - 4 n + 5 n - 5 = 2 n^{2} + n - 3 n t h t e r m (T_{n}) = S_{n} - S_{n - 1} = (2 n^{2} + 5 n) - (2 n^{2} + n - 3) = 4 n + 3$
Question 14 of 25

14. Question
1 points
$I f t h e s u m o f n t e r m s o f a n A P b e 3 n^{2} - n a n d d i s 6, t h e n i t s f i r s t t e r m i s$
- 2
- 3
- 4
- 5
Correct

$S_{n} = 3 n^{2} - n w h e r e n = 1 S_{1} = 3 (1)^{2} - 1 = 3 - 1 = 2 f i r s t t e r m = S_{1} = 2$

Incorrect

$S_{n} = 3 n^{2} - n w h e r e n = 1 S_{1} = 3 (1)^{2} - 1 = 3 - 1 = 2 f i r s t t e r m = S_{1} = 2$
Question 15 of 25

15. Question
1 points
If the sum of the 4th term and the 12th term of an AP is 8,what is the sum of the first 15 terms of the progression ?
- 60
- 110
- 120
- 150
Correct

$l e t t h e f i r s t t e r m a n d c o m m o n d i f f e r e n c e o f A P b e a a n d d T_{4} + T_{12} = 8 a + 3 d + a + 11 d = 8 2 a + 11 d = 8 S_{15} = 15 / 2 [2 a + (15 - 1) d] = 15 / 2 [2 a + 14 d] = 15 / 2 x 8 = 60$

Incorrect

$l e t t h e f i r s t t e r m a n d c o m m o n d i f f e r e n c e o f A P b e a a n d d T_{4} + T_{12} = 8 a + 3 d + a + 11 d = 8 2 a + 11 d = 8 S_{15} = 15 / 2 [2 a + (15 - 1) d] = 15 / 2 [2 a + 14 d] = 15 / 2 x 8 = 60$
Question 16 of 25

16. Question
1 points
The 4th term of 3 times the first and the 7th term exceeds the third term by 1.find the first term a and the common difference d
- a=3,d=2
- a=4,d=3
- a=5,d=4
- a=6d=5
Correct

Incorrect
Question 17 of 25

17. Question
1 points
If n AM are inserted between 7 and 71 and 5th AM is 27 ,then n equal to
- 15
- 16
- 17
- 18
Correct

$i f n A M a r e i n s e r t e d b e t w e e n 7 a n d 71 a r e A_{1} a n d A_{2} . . . . . . . . . . . . . A n 71 A_{1}, A_{2}, A_{3}, . . . . . . . . . A n 71 d = 1 - a / n + 1 = 71 - 7 / n + 1 = 64 / n + 1 g i v e n, A_{5} = 27 a + 5 d = 27 7 + 5 x 64 / n + 1 = 27 320 / n + 1 = 20 n + 1 = 320 / 20 n + 1 = 16 n = 15$

Incorrect

$i f n A M a r e i n s e r t e d b e t w e e n 7 a n d 71 a r e A_{1} a n d A_{2} . . . . . . . . . . . . . A n 71 A_{1}, A_{2}, A_{3}, . . . . . . . . . A n 71 d = 1 - a / n + 1 = 71 - 7 / n + 1 = 64 / n + 1 g i v e n, A_{5} = 27 a + 5 d = 27 7 + 5 x 64 / n + 1 = 27 320 / n + 1 = 20 n + 1 = 320 / 20 n + 1 = 16 n = 15$
Question 18 of 25

18. Question
1 points
In a GP the 6th term is 729 and the common difference is 3 then the first term of GP is
- 2
- 3
- 4
- 5
Correct

$l l l e t t h e l a s t t e r m o f G P i s a a a n d t h e c o m m o n r a t i o i s r g i v e n T_{6} = 729 w h e r e r = 3 a r^{6 - 1} = 729 a (3)^{5} = 729 a x 243 = 729 a = 3$

Incorrect

$l l l e t t h e l a s t t e r m o f G P i s a a a n d t h e c o m m o n r a t i o i s r g i v e n T_{6} = 729 w h e r e r = 3 a r^{6 - 1} = 729 a (3)^{5} = 729 a x 243 = 729 a = 3$
Question 19 of 25

19. Question
1 points
An AP has 13 terns whose sum is 143.the 3rd term is 5 so the 1st term is
- 4
- 7
- 9
- 2
Correct

$g i v e n n = 13, S_{13} = 143, T_{3} = 5, a = ? T_{3} = a + 2 d = 5 . . . . . . . . . . . . (1) S_{n} = \frac{n}{2} [2 a + (n - 1) d] S_{13} = 13 / 2 [2 a + (13 - 1) d] = 143 13 / 2 [2 a + 12 d] = 143 13 / 2 x 2 [a + 6 d] = 143 a + 6 d = 143 / 13 . . . . . . . . . . . . . . . . (2) s o l v i n g 1 & 2 w e g e t a = 2$

Incorrect

$g i v e n n = 13, S_{13} = 143, T_{3} = 5, a = ? T_{3} = a + 2 d = 5 . . . . . . . . . . . . (1) S_{n} = \frac{n}{2} [2 a + (n - 1) d] S_{13} = 13 / 2 [2 a + (13 - 1) d] = 143 13 / 2 [2 a + 12 d] = 143 13 / 2 x 2 [a + 6 d] = 143 a + 6 d = 143 / 13 . . . . . . . . . . . . . . . . (2) s o l v i n g 1 & 2 w e g e t a = 2$
Question 20 of 25

20. Question
1 points
$I f G M o f (a, b, c, d)^{1 / 4} = 3, t h e n G M o f \frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d} w i l l b e$
- 1/3
- 3
- 81
- 1/81
Correct

$G M o f (a, b, c, d)^{1 / 4} = 3 a . b . c . d = 3^{4} a . b . c . d = 81, t h e n G M o f \frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d} = [\frac{1}{a} . \frac{1}{b} . \frac{1}{c} . \frac{1}{d}]^{1 / 4} = [\frac{1}{a b c d}]^{1 / 4} = (a b c d)^{- 1 x 1 / 4} = (81)^{- 1 / 4} = 3^{4 x - 1 / 4} = 3^{- 1} = 1 / 3$

Incorrect

$G M o f (a, b, c, d)^{1 / 4} = 3 a . b . c . d = 3^{4} a . b . c . d = 81, t h e n G M o f \frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d} = [\frac{1}{a} . \frac{1}{b} . \frac{1}{c} . \frac{1}{d}]^{1 / 4} = [\frac{1}{a b c d}]^{1 / 4} = (a b c d)^{- 1 x 1 / 4} = (81)^{- 1 / 4} = 3^{4 x - 1 / 4} = 3^{- 1} = 1 / 3$
Question 21 of 25

21. Question
1 points
$T h e s u m o f t h e i n f i n i t y G P 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + . . . . . . . . . . . . \infty =$
- 1.95
- 1.5
- 1.75
- none of these
Correct

<
$1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + . . . . . . . . . . . . \infty =$
a=1,r=1/3/1=1/3<1

sum=a/1-r=1/1-1/3=1/2/3=3/2=1.5

Incorrect

<
$1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + . . . . . . . . . . . . \infty =$
a=1,r=1/3/1=1/3<1

sum=a/1-r=1/1-1/3=1/2/3=3/2=1.5
Question 22 of 25

22. Question
1 points
$I f x, y, z a r e t h e t e r m s i n G P t h e n t h e t e r m s x^{2} + y^{2}, x y + y z, y^{2} + z^{2} a r e i n$
- AP
- GP
- HP
- none of these
Correct

$I f x, y, z a r e t h e t e r m s i n G P t h e n x = 1, y = 2, z = 4 x^{2} + y^{2}, x y + y z, y^{2} + z^{2} = 1^{2} + 2^{2}, 1 x 2 + 2 x 4, 2^{2} + 4^{2} = 5, 10, 20 a r e i n G P$

Incorrect

$I f x, y, z a r e t h e t e r m s i n G P t h e n x = 1, y = 2, z = 4 x^{2} + y^{2}, x y + y z, y^{2} + z^{2} = 1^{2} + 2^{2}, 1 x 2 + 2 x 4, 2^{2} + 4^{2} = 5, 10, 20 a r e i n G P$
Question 23 of 25

23. Question
1 points
$I f t h e s u m o f f i r s t n t e r m s a r e i n A P i s 6 n^{2} + 6 n, t h e n t h e 4 t h t e r m o f t h e s e r i e s$
- 120
- 72
- 48
- 24
Correct

Incorrect
Question 24 of 25

24. Question
1 points
If a,b,c are in AP then the value of a-b+c is
- a
- -b
- b
- c
Correct

If a,b,c are in AP

then a=1,b=2,c=3

a-b+c=1-2+3=-1+3=2=b

Incorrect

If a,b,c are in AP

then a=1,b=2,c=3

a-b+c=1-2+3=-1+3=2=b
Question 25 of 25

25. Question
1 points
Find the 2 numbers whose GM is 5 and AM is 7.5
- 10 and 5
- 13.09 and 1.91
- 12 and 3
- none of these
Correct

$l e t t h e 2 n u m b e r s b e a a n d b A M = a + b / 2 = 7.5 a + b = 15 . . . . . . . (1) G M = \sqrt{a b = 5} a b = 25 . . . . . . . . . (2) s o l v i n g 1 & 2 w e g e t a = 13.09 a n d b = 1.91$

Incorrect

$l e t t h e 2 n u m b e r s b e a a n d b A M = a + b / 2 = 7.5 a + b = 15 . . . . . . . (1) G M = \sqrt{a b = 5} a b = 25 . . . . . . . . . (2) s o l v i n g 1 & 2 w e g e t a = 13.09 a n d b = 1.91$

Sequence and Series – Test 2

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