## 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 2018.** JEE Main 2018 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 one-fourth of the next five terms. The 20

^{th}term isCorrectLet 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

IncorrectLet 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

CorrectCommon terms are 5, 11, 17, ….

∴

*n*^{th}term of common sequencet

_{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)

IncorrectCommon terms are 5, 11, 17, ….

∴

*n*^{th}term of common sequencet

_{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

*p*th,*q*th,*r*th terms of GP are*a, b*and c respectively, then a^{q-r}b^{r-p}c^{p-q}is equal toCorrectLet the first term of GP is A and common ratio is R.

Given, pth term = T

_{p}= a ⇒ AR^{p-1}= a …….(i)qth term = T

_{q}= b ⇒ AR^{q}^{ – 1}= b …….(ii)rth term = T

_{r}= c ⇒ AR^{r-1}= c …….(iii)∴ a

^{q-r}b^{r}^{-p}c^{p}^{-a}= (AR

^{p-1})^{q-r }(AP^{q-1})r-p (AR^{r-1})^{p-q}= A

^{q-r}R^{(p-1)(q-r)}A^{r-p}R^{(q-1)(r-p)}A^{p}^{-q}R^{(r-1)(p-q)}= A

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

^{0}R^{pq-pr}^{-q+r+qr-pq+r+p+rp-rq-p+q}= A

^{0}R^{0}= 1IncorrectLet the first term of GP is A and common ratio is R.

Given, pth term = T

_{p}= a ⇒ AR^{p-1}= a …….(i)qth term = T

_{q}= b ⇒ AR^{q}^{ – 1}= b …….(ii)rth term = T

_{r}= c ⇒ AR^{r-1}= c …….(iii)∴ a

^{q-r}b^{r}^{-p}c^{p}^{-a}= (AR

^{p-1})^{q-r }(AP^{q-1})r-p (AR^{r-1})^{p-q}= A

^{q-r}R^{(p-1)(q-r)}A^{r-p}R^{(q-1)(r-p)}A^{p}^{-q}R^{(r-1)(p-q)}= A

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

^{0}R^{pq-pr}^{-q+r+qr-pq+r+p+rp-rq-p+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*, then the numbers are^{2}= 27CorrectH = 4, G

^{2}= AH⇒ 2A + G

^{2}= 27⇒ 2A + AH = 27

⇒ 6A = 2A ⇒ A = 9/2, G

^{2}= 18a + b = 9, ab = 18

∴ a = 3, b = 6 or a = 6, b = 3

IncorrectH = 4, G

^{2}= AH⇒ 2A + G

^{2}= 27⇒ 2A + AH = 27

⇒ 6A = 2A ⇒ A = 9/2, G

^{2}= 18a + 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 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.

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 inCorrectLet 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.

IncorrectLet 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

CorrectHere, number of factors = 50

∴The coefficient of x

^{49}= -1 – 3 – 5 – … – 99= -2500

IncorrectHere, 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

CorrectSuppose 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

IncorrectSuppose 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

Correctt

_{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 = 521Incorrectt

_{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

CorrectSuppose 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°.

IncorrectSuppose 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

Correct28, A1, A2 … …. A10, A11, 10

A

_{1}= 28 + d ⇒ clearly not integralThe 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

Incorrect28, A1, A2 … …. A10, A11, 10

A

_{1}= 28 + d ⇒ clearly not integralThe 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 isCorrectSince, 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)

IncorrectSince, 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 beCorrectLet a, ar, ar

^{2}are in GP and a, ar, ar^{2}– 64 are in AP, we geta(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.

IncorrectLet a, ar, ar

^{2}are in GP and a, ar, ar^{2}– 64 are in AP, we geta(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) = 81547 = 7.11

^{2}→ number of divisors = (1 + 1) (1 +3) = 6122 = 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) = 81547 = 7.11

^{2}→ number of divisors = (1 + 1) (1 +3) = 6122 = 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 isCorrectIncorrect - Question 16 of 20
##### 16. Question

The sum of the integers from 1 to 100 which are not divisible by 3 or 5 is

CorrectIncorrect - 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 toCorrectIncorrect - Question 18 of 20
##### 18. Question

The sum of (x + 2)

^{n-1}+ (x + 2)^{n-2}(x + 1) + (x + 2)^{n-3}(x + 1)^{2}+ … + (x + 1)^{n – 1 }is equal toCorrectIncorrect - 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?CorrectIncorrect - 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 isCorrectIncorrect