Test: Sequence and Series - JEE MCQ

The given series is: 1,2,4,7,11,...
Difference between second and first term = 2 - 1 = 1
Difference between third and second term = 4 - 2 = 2
Difference between fourth and third term = 7 - 4 = 3
Difference between fifth and fourth term = 11 - 7 = 4
Difference between sixth and fifth term = 16 - 11 = 5

The given sequence is based on the following pattern:

∴ Required number = 30.

a = 8−6i 
d = 7−4i−8+6i
= −1+2i
an = a+(n−1)d
a+ib = 8−6i+(n−1)(−1+2i)
a+ib = 8−6i+(−1)(n−1)+(n−1)2i
= − 6+2(n−1)=0
= 2(n−1) = 6
n = 4
an = 8−6i+(4−1)(−1+2i)
= 8−6i−3+6i = 5
4th term = 5

Given that a, 4, b form an arithmetic progression (A.P.) and a, 2, b form a geometric progression (G.P.), we need to determine the type of progression formed by a, 1, b. 1. From A.P.: Since a, 4, b are in A.P., the common difference gives: 4 - a = b - 4 \implies a + b = 8 \quad (Equation 1) 2. From G.P.: Since a, 2, b are in G.P., the common ratio gives: \frac{2}{a} = \frac{b}{2} \implies ab = 4 \quad (Equation 2) 3. Solving Equations 1 and 2: From Equation 1, b = 8 - a. Substitute into Equation 2: a(8 - a) = 4 \implies a² - 8a + 4 = 0 Solving this quadratic equation using the quadratic formula a = \frac{-b \pm \sqrt{b² - 4ac}}{2a}: a = \frac{8 \pm \sqrt{64 - 16}}{2} = \frac{8 \pm 4\sqrt{3}}{2} = 4 \pm 2\sqrt{3} Thus, a = 4 + 2\sqrt{3} and b = 4 - 2\sqrt{3}, or vice versa. 4. Checking Progression Type for a, 1, b: - A.P. Check: 1 - a \neq b - 1 Hence, not an A.P. - G.P. Check: \frac{1}{a} \neq \frac{b}{1} Hence, not a G.P. - H.P. Check (Reciprocals): Compute reciprocals: \frac{1}{a}, 1, \frac{1}{b} Simplifying \frac{1}{4 + 2\sqrt{3}} = \frac{2 - \sqrt{3}}{2} and \frac{1}{4 - 2\sqrt{3}} = \frac{2 + \sqrt{3}}{2}. The reciprocals form an A.P. with common difference \frac{\sqrt{3}}{2}. Since the reciprocals are in A.P., a, 1, b are in harmonic progression (H.P.). Thus, the correct answer is B.

The sequence is the Fibonacci series
1+1 = 0
1+2 = 3
2+3 = 5
3+5 = 8
5+8 = 13
8+13 = 21
13+21 = 34
21+34 = 55
34+55 = 89
The 11th term will be 89.

S_p = Sq
⇒  p/2(2a+(p−1)d) = q/2(2a+(q−1)d)
⇒ p(2a+(p−1)d) = q(2a+(q−1)d)
⇒ 2ap + p²d − pd = 2aq + q²d − qd
⇒ 2a(p−q) + (p+q)(p−q)d − d(p−q) = 0
⇒ (p−q)[2a + (p+q)d − d] = 0
⇒ 2a + (p+q)d − d = 0
⇒ 2a + ((p+q) − 1)d = 0
⇒ S_p+q = 0

Let A1, A2, A3, ........ , A40 be 40 A.M's between two numbers 'a' and 'b'.
Then, 
a, A1, A2, A3, ........ , A40, b is an A.P. with common difference d  = (b - a)/(n + 1) = (b - a)/41
[ where n = 40]
now, A1, A2, A3, ........ , A40  = 40/2( A1 + A40)
A1, A2, A3, ........ , A40 = 40/2(a + b)
[ a, A1, A2, A3, ........ , A40, b is an Ap then ,a + b = A1 + A40]
sum of 40A.M = 120(given)
120= 20(a + b)
=> 6 = a + b ----------(1)
Again, consider B1, B2, ........ , B50  be 50 A.M.'s between two numbers a and b.
Then, a, B1, B2, ........ , B50, b will be in A.P. with common difference = ( b - a)/51
now , similarly,
B1, B2, ........ , B50 = 50/2(B1 + B2)
= 25(6) ----------------from(1)
= 150

Let the first term of the given AP be ‘a' and the common difference be ‘d'. Then, the sum of first ‘n' terms of the AP is given by:
Sn =  n/2 {2a+(n-1)d} …….(1)
Here, it is given that:
Sp = q and Sq = p
Using (1), we get:-
q = p/2 {2a+(p-1)d}
and p = q/2 {2a+(q-1)d}
i.e. 2a + (p-1)d = 2q/p …..(2)
and 2a + (q-1)d = 2p/q …..(3)
Subtracting (3) from (2), we get:
(p - 1 - q + 1)d = 2q/p - 2p/q
So, d = 2(q2 - p2)/pq(p-q)
i.e. d = -2(p+q)/pq
Now, substituting the value of ‘d' in eq.n (2), we get:
2a + (p-1){-2(p+q)/pq} = 2q/p
i.e. 2a= 2q/p + 2(p-1)(p+q)/pq
This gives:
a = (p2 + q2 - p - q + pq)/pq
So, we have
Sp+q = (p+q)/2 { 2(p²+q²-p-q+pq)/pq - (p+q-1) 2 (p+q)/pq}
i.e. Sp+q = (p+q)/pq { p²+q²-p-q+pq-p²-pq-qp-q²+p+q}
So, Sp+q = -(p+q)

Given: - The p-th term of an A.P. is q. - The q-th term of the same A.P. is p. We use the formula for the n-th term of an A.P.: a_n = a₁ + (n - 1)d From the given information, we have two equations: 1. q = a₁ + (p - 1)d 2. p = a₁ + (q - 1)d Subtracting equation 2 from equation 1: q - p = (p - q)d This simplifies to: (q - p)(1 + d) = 0 Thus, either: - q = p, which is trivial and doesn't provide new information. - 1 + d = 0 ⇒ d = -1. Assuming d = -1, substitute back into equation 1: q = a₁ - (p - 1) ⇒ a₁ = p + q - 1 Now, find the (p+q)-th term: a_{p+q} = a₁ + (p + q - 1)d = (p + q - 1) + (p + q - 1)(-1) = (p + q - 1) - (p + q - 1) = 0 Thus, the (p+q)-th term is 0.

Given pth term of HP = qr
So pth term of AP = 1/qr
a+(p−1)d = 1/qr....(1)
and qth term of HP = pr
so qth term of AP = 1/pr
a + (q−1)d = 1/pr.....(2)
subtracting equation 1 and 2 we get, 
(p−q)d = (p−q)/pqr
d = 1/pqr
Now from equation 1, 
a = 1/qr − (p−1)/pqr
= (p−p+1)/pqr = 1/pqr
So rth term of AP = a+(r−1)d = 1/pqr + (r − 1)/pqr = 1/pq 
So, rth term of HP = pq

Between n & n², numbers divisible by n are:
2n, 3n, 4n, ….... (n – 1)n
No. of numbers = (n – 1) – 2 + 1 = n – 2

 First A.P′s sequence is 3,7,11,....,407
General term will be 4k+3    k≤101
Second A.P's sequence is 2,9,16,....,709
General term will be 7p+2    p≤101
The common terms will be 51,79,...,28m + 51
28m + 51 ≤ 407
⟹ 28k ≤ 356
⟹ k ≤ 12.71
And adding the 2 starting number count of the 2 A.P′s i.e. 3 & 2 
Number of common terms will be 12 + 2 = 14

Given that a, b, and c are in both an arithmetic progression (A.P.) and a geometric progression (G.P.), we analyze the conditions: 1. Arithmetic Progression (A.P.): The terms satisfy 2b = a + c. 2. Geometric Progression (G.P.): The terms satisfy b² = ac. Substituting a = 2b - c from the A.P. condition into the G.P. equation: b² = (2b - c)c implies b² = 2bc - c². Rearranging gives: c² - 2bc + b² = 0 implies (c - b)² = 0 implies c = b. Substituting c = b back into the A.P. condition: a = 2b - b = b. Thus, a = b = c, proving that all three terms must be equal for them to form both an A.P. and a G.P.

Each term increases by consecutive integers. Starting from 1, add 2 to get 3, add 3 to get 6, add 4 to get 10, and adding 5 gives 15. This is also the fifth triangular number (5^×6/2=15).

For any two unequal positive numbers, the Arithmetic Mean (A.M.) is always greater than the Geometric Mean (G.M.), which in turn is greater than the Harmonic Mean (H.M.). Therefore, the correct order is H < G < A.

take lcm
a+b-2√ab
    2
(√a - √b )^{2
       2}
 

 Since a,b,c are in H.P, so b = 2ac/(a+c).
Also, b,c,d are in H.P, so c = 2bd/(b+d).
Therefore, (a+c)(b+d) = 2ac/b × 2bd/c
⇒ab+cb+ad+cd = 4ad
⇒ab+bc+cd = 3ad

The 2-digit number which when divided by 3 gives remainder 1 are: 10, 13, 16, ...97
Here a = 10, d = 13 - 10 = 3
t_n = 97
nth term of an AP is t_n = a + (n – 1)d
97 = 10 + (n – 1)³
⇒ 97 = 10 + 3n – 3
⇒ 97 = 7 + 3n
⇒ 3n = 97 – 7 = 90
∴ n = 90/3 = 30
Recall sum of n terms of AP, 
= 15[20 + 87] = 15 × 107 = 1605

Clearly, the numbers between 105 and 1000 which are divisible by 7 are 112,119,126,...,994.

This is an AP with first term a=112, common difference d=7 and last term  l = 994.
Let there be n terms in this AP. Then,
an​ = 994

⇒ a + (n − 1)d

994 = 112 + (n − 1) × 7

994 = 105 + 7n

∴ 7n = 889

∴ n = 127