JEE Main Mathematics Sequences and Series Sequences Online Test
JEE Main Mathematics Sequences and Series Sequences Online Test
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JEE Main Mathematics Sequences and Series Sequences Online Test. JEE Main Online Test for Mathematics Sequences and Series Sequences. JEE Main Full Online Quiz for Mathematics Sequences and Series Sequences. JEE Main Free Mock Test Paper 2021. JEE Main 2021 Free Online Practice Test, Take JEE Online Test for All Subjects. JEE Main Question and Answers for Mathematics Sequences and Series Sequences. In this test You may find JEE Main all subjects Questions with Answers. Check JEE Main Question and Answers in English. This mock Test is free for All Students. Mock Test Papers are very helpful for Exam Purpose, by using below Mock Test Paper you may Test your Study for Next Upcoming Exams. Now Scroll down below n Take JEE Main Mathematics Sequences and Series Sequences…
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Question 1 of 20
1. Question
In an AP, the first term is 2 and the sum of the first five terms is onefourth of the next five terms. The 20^{th}term is
Correct
Let AP series is a, a + d, a + 2d, a + 3d, ……
Now, given a = 2
According to the given condition,
Sum of first five terms = 1/4 (Sum of next five terms)
⇒ a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d)
= 1/4 [a + 5d + a + 6d + a + 7d + a + 8d + a + 9d]
⇒ 5a + 10d = 1/4 (5a + 35d)
⇒ 4(5a + 10d) = 5a + 35d
⇒ 20a + 40d = 5a + 35d ⇒ 20a – 5a = 35d – 40d
⇒ 15a = 5d
⇒ 15 × 2 = 5d
⇒ 30 5d (∵ a = 2)
Now, T_{n} = a + (n – 1)d
⇒ T_{20} = 2 + (20 – 1)(6) = 2 + 19 (6)
= 2 – 19 × 6 = 2 – 114 = 112
Incorrect
Let AP series is a, a + d, a + 2d, a + 3d, ……
Now, given a = 2
According to the given condition,
Sum of first five terms = 1/4 (Sum of next five terms)
⇒ a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d)
= 1/4 [a + 5d + a + 6d + a + 7d + a + 8d + a + 9d]
⇒ 5a + 10d = 1/4 (5a + 35d)
⇒ 4(5a + 10d) = 5a + 35d
⇒ 20a + 40d = 5a + 35d ⇒ 20a – 5a = 35d – 40d
⇒ 15a = 5d
⇒ 15 × 2 = 5d
⇒ 30 5d (∵ a = 2)
Now, T_{n} = a + (n – 1)d
⇒ T_{20} = 2 + (20 – 1)(6) = 2 + 19 (6)
= 2 – 19 × 6 = 2 – 114 = 112

Question 2 of 20
2. Question
Number of identical terms in the sequence 2, 5, 8, 11, ……upto 100 terms and 3, 5, 7, 9, 11, …. upto 100 terms are
Correct
Common terms are 5, 11, 17, ….
∴ n^{th} term of common sequence
t_{n} = 5 + (n – 1)6 = (6n – 1)
Also 100^{th} terms of the first sequence
= 2 + (100 – 1)3 = 299
and 100^{th} term of the second sequence
= 3 + (100 – 1)2 = 201
⇒ t_{n} ≤ 201
⇒ 6n – 1 ≤ 201
⇒ n ≤ 33
∴ n = 33 (∵ n ∈ N)
Incorrect
Common terms are 5, 11, 17, ….
∴ n^{th} term of common sequence
t_{n} = 5 + (n – 1)6 = (6n – 1)
Also 100^{th} terms of the first sequence
= 2 + (100 – 1)3 = 299
and 100^{th} term of the second sequence
= 3 + (100 – 1)2 = 201
⇒ t_{n} ≤ 201
⇒ 6n – 1 ≤ 201
⇒ n ≤ 33
∴ n = 33 (∵ n ∈ N)

Question 3 of 20
3. Question
If the pth, qth, rth terms of GP are a, b and c respectively, then a^{qr} b^{rp} c^{pq} is equal to
Correct
Let the first term of GP is A and common ratio is R.
Given, pth term = T_{p} = a ⇒ AR^{p1} = a …….(i)
qth term = T_{q} = b ⇒ AR^{q}^{ – 1} = b …….(ii)
rth term = T_{r} = c ⇒ AR^{r1} = c …….(iii)
∴ a^{qr} b^{r}^{p} c^{p}^{a}
= (AR^{p1})^{qr }(AP^{q1})rp (AR^{r1})^{pq}
= A^{qr}R^{(p1)(qr)}A^{rp} R^{(q1)(rp)} A^{p}^{q} R^{(r1)(pq)}
= A^{q}^{r+rp+pq} R^{(p1)(qr) + (q1)(rp) + (r1)(pq)}
= A^{0} R^{pqpr}^{q+r+qrpq+r+p+rprqp+q}
= A^{0} R^{0} = 1
Incorrect
Let the first term of GP is A and common ratio is R.
Given, pth term = T_{p} = a ⇒ AR^{p1} = a …….(i)
qth term = T_{q} = b ⇒ AR^{q}^{ – 1} = b …….(ii)
rth term = T_{r} = c ⇒ AR^{r1} = c …….(iii)
∴ a^{qr} b^{r}^{p} c^{p}^{a}
= (AR^{p1})^{qr }(AP^{q1})rp (AR^{r1})^{pq}
= A^{qr}R^{(p1)(qr)}A^{rp} R^{(q1)(rp)} A^{p}^{q} R^{(r1)(pq)}
= A^{q}^{r+rp+pq} R^{(p1)(qr) + (q1)(rp) + (r1)(pq)}
= A^{0} R^{pqpr}^{q+r+qrpq+r+p+rprqp+q}
= A^{0} R^{0} = 1

Question 4 of 20
4. Question
The HM of two numbers is 4 and their AM and GM satisfy the relation 2A + G^{2} = 27, then the numbers are
Correct
H = 4, G^{2} = AH
⇒ 2A + G^{2} = 27
⇒ 2A + AH = 27
⇒ 6A = 2A ⇒ A = 9/2, G^{2} = 18
a + b = 9, ab = 18
∴ a = 3, b = 6 or a = 6, b = 3
Incorrect
H = 4, G^{2} = AH
⇒ 2A + G^{2} = 27
⇒ 2A + AH = 27
⇒ 6A = 2A ⇒ A = 9/2, G^{2} = 18
a + b = 9, ab = 18
∴ a = 3, b = 6 or a = 6, b = 3

Question 5 of 20
5. Question
If a, b, c and d are in GP, then
a^{n} + b^{n}, b^{n} + c^{n}, c^{n} + d^{n} are in
Correct
∵ a, b, c, d are in G.P.
⇒ b = ar, c = ar^{2}, d = ar^{2} ……..(i)
Now, (a^{n }+ b^{n}) (c^{n} + d^{n})
= (a^{n} + a^{n}r^{n}) × (a^{n}r^{2n} + a^{n}r^{3n}) [from Eq. (i)]
= a^{n} (1 + rn) a^{n}r^{2n} (1 + r^{n})
= a^{2n}r^{2n}(1 + r^{n})^{2} = (a^{n}r^{n}(1 + r^{n}))2 = [a^{n}r^{n} + a^{n}r^{2n}]^{2}
= [(ar)^{n} + (ar^{2})^{n}]^{2} = (b^{n} + c^{n})^{2} [from Eq. (i)]
Hence, they are in GP.
Incorrect
∵ a, b, c, d are in G.P.
⇒ b = ar, c = ar^{2}, d = ar^{2} ……..(i)
Now, (a^{n }+ b^{n}) (c^{n} + d^{n})
= (a^{n} + a^{n}r^{n}) × (a^{n}r^{2n} + a^{n}r^{3n}) [from Eq. (i)]
= a^{n} (1 + rn) a^{n}r^{2n} (1 + r^{n})
= a^{2n}r^{2n}(1 + r^{n})^{2} = (a^{n}r^{n}(1 + r^{n}))2 = [a^{n}r^{n} + a^{n}r^{2n}]^{2}
= [(ar)^{n} + (ar^{2})^{n}]^{2} = (b^{n} + c^{n})^{2} [from Eq. (i)]
Hence, they are in GP.

Question 6 of 20
6. Question
If a^{1}^{/x} = b^{1}^{/y} = c^{1}^{/z} and a,b,c ae in GP, then x,y,z will be in
Correct
Let a^{1}^{/x} = b^{1}^{/y} = c^{1}^{/z}
= k ⇒ a = k^{x}, b = k^{y}, c = k^{z}
Now, a,b,c are in G.P.
⇒ b^{2} = ac ⇒ k^{2y} = k^{x}.k^{z} = k^{x}^{ + z}
⇒ 2y = x + z
∴ x,y,z are in A.P.
Incorrect
Let a^{1}^{/x} = b^{1}^{/y} = c^{1}^{/z}
= k ⇒ a = k^{x}, b = k^{y}, c = k^{z}
Now, a,b,c are in G.P.
⇒ b^{2} = ac ⇒ k^{2y} = k^{x}.k^{z} = k^{x}^{ + z}
⇒ 2y = x + z
∴ x,y,z are in A.P.

Question 7 of 20
7. Question
The coefficient of x^{49} in the product
(x – 1)(x – 3) … .. (x – 99) is
Correct
Here, number of factors = 50
∴The coefficient of x^{49} = 1 – 3 – 5 – … – 99
= 2500
Incorrect
Here, number of factors = 50
∴The coefficient of x^{49} = 1 – 3 – 5 – … – 99
= 2500

Question 8 of 20
8. Question
The number which should be added to the numbers 2, 14, 62, so that the resulting numbers may be in GP, is
Correct
Suppose that the added number be x, then x + 2, x + 14, x + 62 are in GP.
∴ (x + 14)^{2} = (x + 2)(x + 62)
⇒ x^{2} + 196 + 28x = x^{2} + 64x + 124
⇒ 36x = 72 ⇒ x = 2
Incorrect
Suppose that the added number be x, then x + 2, x + 14, x + 62 are in GP.
∴ (x + 14)^{2} = (x + 2)(x + 62)
⇒ x^{2} + 196 + 28x = x^{2} + 64x + 124
⇒ 36x = 72 ⇒ x = 2

Question 9 of 20
9. Question
The largest term common to the sequences
1, 11, 21, 31, …… to 100 terms and 31, 36, 41, 46, ….. to 100 terms
Correct
t_{m} + t_{n}
1 + (m – 1), 10 = 31 + (n – 1).5
10m – 9 = 5n + 26
10m = 5n + 34 = 5(n + 7)
m = λ and n = 2λ – 7
m ≤ 100 and n ≤ 100
λ ≤ 100 and 2λ – 7 ≤ 100 ⇒ λ ≤ 53 1/2
so 1 ≤ λ ≤ 53 [No. of common term = 53]
t_{53} = 1 + 52.10 = 521
Incorrect
t_{m} + t_{n}
1 + (m – 1), 10 = 31 + (n – 1).5
10m – 9 = 5n + 26
10m = 5n + 34 = 5(n + 7)
m = λ and n = 2λ – 7
m ≤ 100 and n ≤ 100
λ ≤ 100 and 2λ – 7 ≤ 100 ⇒ λ ≤ 53 1/2
so 1 ≤ λ ≤ 53 [No. of common term = 53]
t_{53} = 1 + 52.10 = 521

Question 10 of 20
10. Question
If the angles of a quadrilateral are in AP, whose common difference is 10°, then the angles of the quadrilateral are
Correct
Suppose that ∠A = x, then ∠B = x + 10^{o},
∠C = x + 20^{o} and ∠D = x + 30^{o}
So, we know that ∠A + ∠B + ∠C + ∠D = 360^{o}
On putting these values, we get
(x) + (x + 10°) + (x + 20°) + (x + 30°) = 360°
⇒ x = 75°
Hence, the angles of the quadrilateral are 75°, 85°, 95°, 105°.
Incorrect
Suppose that ∠A = x, then ∠B = x + 10^{o},
∠C = x + 20^{o} and ∠D = x + 30^{o}
So, we know that ∠A + ∠B + ∠C + ∠D = 360^{o}
On putting these values, we get
(x) + (x + 10°) + (x + 20°) + (x + 30°) = 360°
⇒ x = 75°
Hence, the angles of the quadrilateral are 75°, 85°, 95°, 105°.

Question 11 of 20
11. Question
If eleven A.M’s are inserted between 28 and 10 then the number of integral A.M.’s
Correct
28, A1, A2 … …. A10, A11, 10
A_{1} = 28 + d ⇒ clearly not integral
The A.M.’s will be integral only when 2d, 4d ………types of terms occur So, A_{2}, A_{4}, A_{6}, A_{8}, A_{10} are the A.M.
Total number = 5
Incorrect
28, A1, A2 … …. A10, A11, 10
A_{1} = 28 + d ⇒ clearly not integral
The A.M.’s will be integral only when 2d, 4d ………types of terms occur So, A_{2}, A_{4}, A_{6}, A_{8}, A_{10} are the A.M.
Total number = 5

Question 12 of 20
12. Question
If x, 2y and 3z are in AP, where the distinct numbers x, y, z are in GP, then the common ratio of the GP is
Correct
Since, x, 2y and 3z are in AP.
∴ 4y = x + 3z …..(i)
And x, y, z are in GP.
∴ y = rx and z = xr^{2}
On putting the values of y and z in Eq. (i), we get 4xr = x + 3xr^{2}
⇒ 3r^{2} – 4r + 1 = 0
⇒ (3r – 1)(r – 1) = 0
⇒ r = 1/3, 1
∴ r = 1/3 (∵ r ≠ 1)
Incorrect
Since, x, 2y and 3z are in AP.
∴ 4y = x + 3z …..(i)
And x, y, z are in GP.
∴ y = rx and z = xr^{2}
On putting the values of y and z in Eq. (i), we get 4xr = x + 3xr^{2}
⇒ 3r^{2} – 4r + 1 = 0
⇒ (3r – 1)(r – 1) = 0
⇒ r = 1/3, 1
∴ r = 1/3 (∵ r ≠ 1)

Question 13 of 20
13. Question
Three numbers form a GP. If the 3^{rd} term is decreased by 64, then the three numbers thus obtained will constitute an AP. If the second term of this AP is decreased by 8, a GP will be formed again, then the numbers will be
Correct
Let a, ar, ar^{2} are in GP and a, ar, ar^{2} – 64 are in AP, we get
a(r^{2} – 2r + 1) = 64 ….(i)
Again, a, ar – 8, ar^{2} – 64 are in GP.
∴ (ar – 8)^{2} = a(ar^{2} – 64)
⇒ a(16r – 64) = 64 ….(ii)
On solving Eqs. (i) and (ii), we get r = 5, a = 4
Thus, required numbers are 4, 20, 100.
Incorrect
Let a, ar, ar^{2} are in GP and a, ar, ar^{2} – 64 are in AP, we get
a(r^{2} – 2r + 1) = 64 ….(i)
Again, a, ar – 8, ar^{2} – 64 are in GP.
∴ (ar – 8)^{2} = a(ar^{2} – 64)
⇒ a(16r – 64) = 64 ….(ii)
On solving Eqs. (i) and (ii), we get r = 5, a = 4
Thus, required numbers are 4, 20, 100.

Question 14 of 20
14. Question
The numbers of divisors of 1029, 1547 and 122 are in
Correct
∵ 1029 = 3.7^{3} → number of divisors = (1 + 1) (1 +3) = 8
1547 = 7.11^{2} → number of divisors = (1 + 1) (1 +3) = 6
122 = 2.61 → number of divisors = (1+ 1 ) (1 + 1) = 4
Also 8, 6, 4 are in AP
Incorrect
∵ 1029 = 3.7^{3} → number of divisors = (1 + 1) (1 +3) = 8
1547 = 7.11^{2} → number of divisors = (1 + 1) (1 +3) = 6
122 = 2.61 → number of divisors = (1+ 1 ) (1 + 1) = 4
Also 8, 6, 4 are in AP

Question 15 of 20
15. Question
The harmonic mean between two numbers is 21/5 their A.M. ‘A’ and G.M. ‘G’ satisfy the relation 3A + G^{2} = 36. Then the sum of the square of the numbers is
Correct
Incorrect

Question 16 of 20
16. Question
The sum of the integers from 1 to 100 which are not divisible by 3 or 5 is
Correct
Incorrect

Question 17 of 20
17. Question
The roots of equation x^{2} + 2(a – 3) x + 9 = 0 lie between – 6 and 1 and 2, h_{1}, h_{2}, …, h_{20}. [a] are in HP, where [a] denotes the integral part of a and 2, a_{1}, a_{2}, ……, a_{20}, [a] are in AP, then a_{3}h_{18} is equal to
Correct
Incorrect

Question 18 of 20
18. Question
The sum of (x + 2)^{n1} + (x + 2)^{n2} (x + 1) + (x + 2)^{n3} (x + 1)^{2} + … + (x + 1)^{n – 1 }is equal to
Correct
Incorrect

Question 19 of 20
19. Question
Let S_{1}, S_{2}…. be squares such that for each n ≥ 1, the length of a side of S_{n} equals the length of a diagonal of S_{n}_{ + 1}. If the length of a side of S_{1} is 10 cm, then for which of the following values of n, the area of S_{n} less than 1 sq cm?
Correct
Incorrect

Question 20 of 20
20. Question
If the sum of a certain n number of terms of the AP, 25, 22, 19, …….. is 116, then the last term is
Correct
Incorrect