Sequences and Series Test - 2

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 + p² d −pd = 2aq + q² d −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