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

Start quiz button

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

sequence and series

Quiz-summary

Information

Results

Categories

ekvijteam@gmail.com

1. Question

2. Question

3. Question

4. Question

5. Question

6. Question

7. Question

8. Question

9. Question

10. Question

11. Question

12. Question

13. Question

14. Question

15. Question

16. Question

17. Question

18. Question

19. Question

20. Question

21. Question

22. Question

23. Question

24. Question

25. Question

Latest Articles

mensuration basic concepts

Atomic structure

Environmental issues

Biodiversity and conservation

Ecosystem

Sequence and Series – Test 1

Basic Concept of Permutations and Combinations-Test 2