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All questions of Progressions (A.P. & G.P.) for JAMB Exam

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After striking the floor, a rubber ball rebounds to 4/5th of the height from which it has fallen. Find the total distance that it travels before coming to rest if it has been gently dropped from a height of 120 metres.
  • a)
    540 m
  • b)
    960 m
  • c)
    1080 m
  • d)
    1020 m
  • e)
    1120 m
Correct answer is option 'C'. Can you explain this answer?

Vikas Choudhury answered
The first drop is 120 metres. After this the ball will rise by 96 metres and fall by 96 metres. This process will continue in the form of infinite GP with common ratio 0.8 and first term 96. The required answer is given by the formula:
a/(1-r) 
Now,
[{120/(1/5)}+{96/(1/5)}] 
= 1080 m.

Find  the  15th  term  of  an  arithmetic  progression  whose  first  term  is  2  and  the  common  difference  is 3
  • a)
    45
  • b)
    38
  • c)
    44
  • d)
    40
Correct answer is option 'C'. Can you explain this answer?

Aisha Gupta answered
Method to Solve :

A ( first term ) :- 2

d ( common difference ) :- 3

n = 15

To find nth term we have formula as

an = a + ( n - 1 )d

a15 = 2 + 14 � 3

a15 = 2 + 42

a15 = 44

Four  angles  of  a  quadrilateral  are  in  G.P.  Whose  common  ratio  is  an  intiger.  Two  of  the  angles  are  acute  while  the  other  two  are  obtuse.   The  measure  of  the  smallest   angle  of  the  quadrilateral  is
  • a)
    12
  • b)
    24
  • c)
    36
  • d)
    48
Correct answer is option 'B'. Can you explain this answer?

Kavya Saxena answered
Let   the  angles  be  a, ar, ar 2, ar 3.
Sum  of  the angles = a ( r 4- 1 ) /r -1 = a ( r 2 + 1 ) ( r + 1 ) = 360
a< 90 , and  ar< 90,  Therefore,  a ( 1 + r ) <  180,  or   ( r 2 + 1 ) > 2
Therefore, r  is  not  equal  to  1.  Trying  for  r  =  2  we  get  a  = 24  Therefore, The  angles  are  24, 48, 96  and  192.

How many terms are there in 20, 25, 30......... 140
  • a)
    22
  • b)
    25
  • c)
    23
  • d)
    24
Correct answer is option 'B'. Can you explain this answer?

Anaya Patel answered
Number of terms = { (1st term - last term) / common difference} + 1
= {(140-20) / 5} + 1
⇒ (120/5) + 1
⇒ 24 + 1 = 25.

The internal angles of a plane polygon are in AP. The smallest angle is 100o and the commondifference is 10o. Find the number of sides of the polygon.
  • a)
    8
  • b)
    9
  • c)
    Either 8 or 9
  • d)
    None of these
Correct answer is option 'C'. Can you explain this answer?

Rashi Nambiar answered
The sum of the interior angles of a polygon are multiples of 180 and are given by (n – 1) × 180
where n is the number of sides of the polygon. Thus, the sum of interior angles of a polygon would
be a member of the series: 180, 360, 540, 720, 900, 1080, 1260
The sum of the series with first term 100 and common difference 10 would keep increasing when
we take more and more terms of the series. In order to see the number of sides of the polygon, we
should get a situation where the sum of the series represented by 100 + 110 + 120… should
become a multiple of 180. The number of sides in the polygon would then be the number of terms
in the series 100, 110, 120 at that point.
If we explore the sums of the series represented by 100 + 110 + 120…
We realize that the sum of the series becomes a multiple of 180 for 8 terms as well as for 9 terms.
It can be seen in: 100 + 110 + 120 + 130 + 140 + 150 + 160 + 170 = 1080
Or 100 + 110 + 120 + 130 + 140 + 150 + 160 + 170 + 180 = 1260.

The sum of 5 numbers in AP is 30 and the sum of their squares is 220. Which of the following is the third term?
  • a)
    5
  • b)
    6
  • c)
    8
  • d)
    9
Correct answer is option 'B'. Can you explain this answer?

Isha Mukherjee answered
Since the sum of 5 numbers in AP is 30, their average would be 6. The average of 5 terms in AP is
also equal to the value of the 3rd term (logic of the middle term of an AP). Hence, the third term’s
value would be 6. Option (b) is correct.

If the nth term of AP is m and the nth term of the same AP is m, then (m + n)th term of AP is
  • a)
    m+n
  • b)
    0
  • c)
    m-n 
  • d)
    -m+n
Correct answer is option 'B'. Can you explain this answer?

Impact Learning answered
  • Tn = m = a + (n - l ) d
  • Tm = n= a + (m - l ) d
  • subtracting them we get d=-1
  • and a=m+n-1
  • Add the two and solve through the options given.
  • then (m + n)th term of AP= a+(m+n-1)d
  • Putting values of a and d in the solution, we get
  •                                         =  m+n-1+(m+n-1)x-1
  •                                         = 0

How many terms are there in the GP 5, 20, 80, 320........... 20480?
  • a)
    5
  • b)
    6
  • c)
    8
  • d)
    9
  • e)
    7
Correct answer is option 'E'. Can you explain this answer?

Sameer Rane answered
Common ratio, r = 20/5 = 4;
Last term or nth term of GP = ar^[n-1].
20480 = 5*[4^(n-1)];
Or, 4^(n-1) = 20480/5 = 4^8;
So, comparing the power, 
Thus, n-1 = 8;
Or, n = 7;
Number of terms = 7.

The sum of the first four terms of an AP is 28 and sum of the first eight terms of the same AP is 88.Find the sum of the first 16 terms of the AP?
  • a)
    346
  • b)
    340
  • c)
    304
  • d)
    268
Correct answer is option 'C'. Can you explain this answer?

Abhay Shah answered
Think like this:
The average of the first 4 terms is 7, while the average of the first 8 terms must be 11.
Now visualize this :
Hence, d = 4/2 = 2 [Note: understand this as a property of an A.P.]
Hence, the average of the 6th and 7th terms = 15 and the average of the 8th and 9th term = 19
But this (19) also represents the average of the 16 term A.P.
Hence, required answer = 16 × 19 = 304.

A group of friends have some money which was in an increasing GP. The total money with the first and the last friend was Rs 66 and the product of the money that the second friend had and that the last but one friend had was Rs 128. If the total money with all of them together was Rs 126, then how many friends were there?
  • a)
    6
  • b)
    5
  • c)
    3
  • d)
    Cannot be determined
Correct answer is option 'A'. Can you explain this answer?

Prasenjit Basu answered
The sum of money with the first and the last friend = 66. This can be used as a hint. Let us assume the first friend was having Rs 2 and the last friend was having Rs 64. So, the money can be in the sequence 2, 4, 8, 16, 32, 64. It satisfies the given conditions. Alternatively, this can be done by using the formula for tn of GP also.

How many terms are there in the AP 20, 25, 30,… 130.
  • a)
    22
  • b)
    23
  • c)
    21
  • d)
    24
Correct answer is option 'B'. Can you explain this answer?

Arnav Rane answered
In order to count the number of terms in the AP, use the short cut:
[(last term – first term)/ common difference] + 1. In this case it would become:
[(130 – 20)/5] +1 = 23. Option (b) is correct.

If a man saves ` 4 more each year than he did the year before and if he saves ` 20 in the first year,after how many years will his savings be more than ` 1000 altogether?
  • a)
    19 years
  • b)
    20 years
  • c)
    21years
  • d)
    18 years
Correct answer is option 'A'. Can you explain this answer?

Akash Sengupta answered
We need the sum of the series 20 + 24 + 28 to cross 1000. Trying out the options, we can see that
in 20 years the sum of his savings would be: 20 + 24 + 28 + … + 96. The sum of this series would
be 20 × 58 =1160. If we remove the 20th year we will get the series for savings for 19 years. The
series would be 20 + 24 + 28 + …. 92. Sum of the series would be 1160 – 96 = 1064. If we
remove the 19th year’s savings the savings would be 1064 – 92 which would go below 1000.
Thus, after 19 years his savings would cross 1000. Option (a) is correct.

What  is  the  sum  of  the  first  15  terms  of  an  A.P  whose  11 th  and   7 th  terms  are  5.25  and  3.25  respectively
  • a)
    56.25  
  • b)
    60
  • c)
    52.5
  • d)
    none of these
Correct answer is option 'A'. Can you explain this answer?

Ishani Rane answered
a +10d  = 5.25, a+6d  = 3.25,  4d  =  2,  d  =  1/4
a +5  =  5.25, a  = 0.25  = 1/4,   s 15 =  15/2 ( 2 * 1/4 +  14 * 1/4 )
=  15/2 (1/2 +14/2 )     =  15/2 *15/2  =225/ 4   =   56.25

A number 20 is divided into four parts that are in AP such that the product of the first and fourth is to the product of the second and third is 2 : 3. Find the largest part.
  • a)
    12
  • b)
    4
  • c)
    8
  • d)
    9
Correct answer is option 'C'. Can you explain this answer?

Dipika Iyer answered
Since the four parts of the number are in AP and their sum is 20, the average of the four parts must
be 5. Looking at the options for the largest part, only the value of 8 fits in, as it leads us to think of
the AP 2, 4, 6, 8. In this case, the ratio of the product of the first and fourth (2 × 8) to the product
of the first and second (4 × 6) are equal. The ratio becomes 2:3.

The mid-points of the adjacent sides of a square are joined. Again the mid-points of the adjacent sides of the newly formed figure are connected and this process is repeated again and again. Calculate the sum of the areas of all such figures given that the diagonal of outermost square is 6√2cm.
  • a)
    35 cm2
  • b)
    44 cm2
  • c)
    72 cm2
  • d)
    58 cm2
Correct answer is option 'C'. Can you explain this answer?

Aarav Sharma answered
Let the side of the outermost square be $s$. Then its diagonal is $s\sqrt{2}=6$, so $s=3\sqrt{2}$.

The first figure formed by joining the midpoints of adjacent sides is a square whose side length is $\frac{s}{\sqrt{2}}=\frac{3\sqrt{2}}{2}$. Its area is $\left(\frac{3\sqrt{2}}{2}\right)^2=\frac{27}{4}$.

The second figure formed by joining the midpoints of adjacent sides of the first figure is a square whose side length is $\frac{s}{2}$. Its area is $\left(\frac{s}{2}\right)^2=\frac{18}{4}=4.5$.

The third figure formed by joining the midpoints of adjacent sides of the second figure is a square whose side length is $\frac{s}{2\sqrt{2}}=\frac{3}{2}$. Its area is $\left(\frac{3}{2}\right)^2=\frac{9}{4}$.

This process continues indefinitely, with each successive figure being a square whose side length is half the previous square's side length. The sum of the areas of all these squares is therefore:

$$\frac{27}{4}+4.5+\frac{9}{4}+\frac{9}{16}+\frac{1}{4}+\frac{1}{64}+\cdots$$

This is an infinite geometric series with first term $\frac{27}{4}$ and common ratio $\frac{1}{8}$. Therefore, the sum is:

$$\frac{\frac{27}{4}}{1-\frac{1}{8}}=\frac{\frac{27}{4}}{\frac{7}{8}}=\frac{27}{4}\cdot\frac{8}{7}=\boxed{12}$$

The 4th and 10th term of an GP are 1/3 and 243 respectively. Find the 2nd term.
  • a)
    3
  • b)
    1
  • c)
    1/27
  • d)
    1/9
Correct answer is option 'C'. Can you explain this answer?

Akash Sengupta answered
[Note: To go forward in a G.P. you multiply by the common ratio, to go backward in a G.P. you
divide by the common ratio.]

Find the lowest number in an AP such that the sum of all the terms is 105 and greatest term is 6 times the least.
  • a)
    5
  • b)
    10
  • c)
    15
  • d)
    (a), (b) & (c)
Correct answer is option 'D'. Can you explain this answer?

Aarav Sharma answered
Approach:
Let the first term of the AP be a and the common difference be d.
Given, Sum of all the terms = 105
Also, greatest term = 6 times the least term
Therefore, the greatest term = a + (n-1)d = 6a
where n is the number of terms in the AP.

Calculation:
1. Sum of all terms in an AP:
Sum of n terms in an AP can be given by the formula:
Sum = (n/2)[2a + (n-1)d]
Here, Sum = 105
105 = (n/2)[2a + (n-1)d]

2. Greatest term is 6 times the least term:
a + (n-1)d = 6a
5a = (n-1)d

3. Substitute the value of d in equation 1:
105 = (n/2)[2a + 5a(n-1)/(n-1)]
105 = (n/2)[(7a-5a+5a(n-1))/(n-1)]
105 = (n/2)[(2a+5a(n-1))/(n-1)]

4. Simplify the equation:
210 = n[2a + 5a(n-1)]
Divide both sides by a:
210/a = n(2 + 5n - 5)
42/a = n(5n-3)

5. Check for values of a:
We need to find the lowest value of a.
From the above equation, we can see that a must be a factor of 42.
Therefore, the possible values of a are 1, 2, 3, 6, 7, 14, 21, 42.

6. Substitute values of a to find n:
For each value of a, we can find the corresponding value of n.
If the value of n is a positive integer, then that value of a is valid.
The values of a and n are as follows:
a = 1, n = 15
a = 2, n = 6
a = 3, n = 3.6 (not valid)
a = 6, n = 2.4 (not valid)
a = 7, n = 2.1 (not valid)
a = 14, n = 1.5 (not valid)
a = 21, n = 1.2 (not valid)
a = 42, n = 0.75 (not valid)

7. Find the common difference:
From the equation, 5a = (n-1)d, we get the common difference as:
d = 5a/(n-1)

8. Check for valid values of d:
If the value of d is positive, then the corresponding value of a is valid.
The values of a, n, and d are as follows:
a = 1, n = 15, d = 1/2 (valid)
a = 2, n = 6, d = 2/5 (not valid)
a = 3, n = 3.6, d = 3 (not valid)
a = 6, n = 2.4, d = 6 (not valid)
a = 7, n = 2.1, d

Find the 1st term of an AP whose 8th and 12th terms are respectively 39 and 59.
  • a)
    5
  • b)
    6
  • c)
    4
  • d)
    3
Correct answer is option 'C'. Can you explain this answer?

Isha Mukherjee answered
Since the 8th and the 12th terms of the AP are given as 39 and 59 respectively, the difference
between the two terms would equal 4 times the common difference. Thus we get 4d = 59 – 39 =
20. This gives us d = 5. Also, the 8th term in the AP is represented by a + 7d, we get:
a + 7d = 39 Æ a + 7 × 5 = 39 Æ a = 4. Option (c) is correct.

How many terms are there in the GP 5, 20, 80, 320,… 20480?
  • a)
    6
  • b)
    5
  • c)
    7
  • d)
    8
Correct answer is option 'C'. Can you explain this answer?

Saanvi Sarkar answered
The series would be 5, 20, 80, 320, 1280, 5120, 20480. Thus, there are a total of 7 terms in the
series. Option (c) is correct.

Each of the series 13 + 15 + 17+…. and 14 + 17 + 20+… is continued to 100 terms. Find howmany terms are identical between the two series.
  • a)
    35
  • b)
    34
  • c)
    32
  • d)
    33
Correct answer is option 'D'. Can you explain this answer?

Gayatri Sarkar answered
The two series till their hundredth terms are 13, 15, 17….211 and 14, 17, 20…311. The common
terms of the series would be given by the series 17, 23, 29….209. The number of terms in this
series of common terms would be 192/6 + 1 = 33. Option (d) is correct.

The sum of three numbers in a GP is 14 and the sum of their squares is 84. Find the largestnumber.
  • a)
    8
  • b)
    6
  • c)
    4
  • d)
    12
Correct answer is option 'A'. Can you explain this answer?

Aarav Sharma answered
To solve this problem, we can use the formulas for the sum of a geometric progression (GP) and the sum of the squares of a GP.

Let's assume that the three numbers in the GP are "a/r", "a", and "ar", where "a" is the first term and "r" is the common ratio.

The sum of the three numbers is given as 14:
a/r + a + ar = 14 ----(1)

The sum of their squares is given as 84:
(a/r)^2 + a^2 + (ar)^2 = 84 ----(2)

Now, let's simplify equations (1) and (2) to solve for "a" and "r".

Simplifying equation (1):
a/r + a + ar = 14
a(1/r + 1 + r) = 14
a(r^2 + r + 1)/r = 14
a(r^2 + r + 1) = 14r ----(3)

Simplifying equation (2):
(a/r)^2 + a^2 + (ar)^2 = 84
a^2/r^2 + a^2 + a^2r^2 = 84
a^2(1/r^2 + 1 + r^2) = 84
a^2(r^2 + 1 + r^4)/r^2 = 84
a^2(r^4 + r^2 + 1) = 84r^2 ----(4)

Now, let's substitute the value of "a" from equation (3) into equation (4).

(a(r^2 + r + 1))^2 = 84r^2
(a^2(r^2 + r + 1)^2) = 84r^2
[(14r)/(r^2 + r + 1)]^2(r^2 + r + 1)^2 = 84r^2
(14r)^2 = 84r^2
196r^2 = 84r^2
112r^2 = 0
r^2 = 0

Since the common ratio "r" cannot be zero, this means that there is no valid solution for this problem. Therefore, the given information is inconsistent and there is no largest number that satisfies both conditions.

Hence, the correct answer would be considered as "None of the above" or "Not possible to determine".

A sum of money kept in a bank amounts to ` 1240 in 4 years and ` 1600 in 10 years at simpleInterest. Find the sum.
  • a)
    ` 800
  • b)
    ` 900
  • c)
    ` 1150
  • d)
    ` 1000
Correct answer is option 'D'. Can you explain this answer?

Debanshi Sarkar answered
The difference between the amounts at the end of 4 years and 10 years will be the simple interest
on the initial capital for 6 years.
Hence, 360/6 = 60 =(simple interest.)
Also, the Simple Interest for 4 years when added to the sum gives 1240 as the amount.
Hence, the original sum must be 1000.

A series in which any term is the sum of the preceding two terms is called a Fibonacci series. The first two terms are given initially and together they determine the entire series. If the difference of the squares of the ninth and the eighth terms of a Fibonacci series is 715 then, what is the 12th term of that series?
  • a)
    157
  • b)
    142
  • c)
    144
  • d)
    Cannot be determined
Correct answer is option 'C'. Can you explain this answer?

Aarav Sharma answered
Given:
- The difference of the squares of the ninth and the eighth terms of a Fibonacci series is 715.

To find:
- The 12th term of that series.

Approach:
- We can use the formula to find the nth term of a Fibonacci series: Fn = Fn-1 + Fn-2
- We can also use the given information to form equations and solve for the required term.

Solution:
Let's assume that the first two terms of the Fibonacci series are a and b, then we can write the series as:
a, b, a+b, 2b+a, 3b+2a, 5b+3a, 8b+5a, 13b+8a, 21b+13a, ...

We are given that the difference of the squares of the ninth and the eighth terms of the series is 715, so we can write the equation as follows:
(21b+13a)^2 - (13b+8a)^2 = 715

Simplifying the equation, we get:
(8a+21b+13a+13b)(8a+21b-13a-13b) = 715
(21a+34b)(-5a+8b) = 715

We can find two factors of 715 whose difference is 13*2*5 = 130, which are 65 and 11:
(21a+34b) = 65 and (-5a+8b) = 11
Solving these equations, we get: a = 2 and b = 1.

Now, we can use the formula to find the 12th term of the series:
F12 = F11 + F10
F10 = F9 + F8
F11 = F10 + F9

Using these equations, we can find that F9 = 34, F8 = 21, F10 = 55, and F11 = 89.
Therefore, F12 = F11 + F10 = 89 + 55 = 144.

Hence, the correct answer is option (c) 144.

 If the roots of x3 - 12x2 + 39x - 28 = 0 are in an AP then their common difference is
  • a)
    ± 1
  • b)
    ± 2
  • c)
    ± 3
  • d)
    ± 4
Correct answer is option 'C'. Can you explain this answer?

Gowri Chakraborty answered
Factorize the equation and we get ( x - 1 ) ( x - 4 ) ( x - 7 )
Explanation:

Given Equation: x³ - 12x² + 39x - 28 = 0

Roots in an AP:
Let the roots of the equation be a - d, a, a + d, where d is the common difference.

Sum of roots: = -b/a
Sum of roots = a - d + a + a + d = 3a

From the equation, we know that sum of roots = -(-12) = 12

Therefore, 3a = 12
a = 4

Product of roots: = -d/a
Product of roots = (a - d)(a)(a + d) = 4(a2 - d2) = 28

Substitute a = 4 in the equation:
42 - d² = 7
16 - d² = 7
d² = 9
d = ±3


Therefore, the correct answer is C: ±3.

Chapter doubts & questions for Progressions (A.P. & G.P.) - Mathematics for JAMB 2025 is part of JAMB exam preparation. The chapters have been prepared according to the JAMB exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for JAMB 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

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