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

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 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

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