Test: Introduction To Sequences - JEE MCQ

a_n = (n²)/2ⁿ
⇒ a₅ = [(5)²]/2⁽⁵⁾
⇒ a₅ = 25/32

a1 = 2 
a2 = 2a1 + 1
=> 2(2) + 1 = 5
a3 = 2a2 + 1
=> 2(5) + 1 = 11
a4 = 2a3 + 1
=> 2(11) + 1 = 23
Hence, the required series is : 2,5,11,23………

an = (n-1)(2-n)(3+n)
Put n = 10
an = 9×(-8)×13
= - 936

Case-I: x<1/3 ⇒ −3x+1,3,−x+3 are in A.P.
⇒ 6 = −3x + 1 − x + 3
⇒ 4x = −2 ⇒ x = −1/2
⇒ terms are 5/2, 3, 7/2, 4, 9/2
⇒ sum =35/2
Case-II: 1/3 ≤ x < 3
3x − 1,3, −x + 3 are in A.P.
⇒ 6 = 3x − 1 − x + 3 = 2x + 2 ⇒ x = 2
⇒ terms are 5, 3, 1, –1, –3
⇒ sum = 5
Case-III: x ≥ 3
⇒ terms are 3x − 1,3,x − 3
⇒ 6 = 3x − 1 + x − 3 = 4x − 4
⇒ 4x = 10 ⇒ x = 5/2 not possible

Given sequence 8A+2B,6A+B,4A,2A−B,……
This is an AP with common difference =−2 A−B
First term, a = 8 A + 2 B
nth  term of AP,tn=a+(n−1)d
= 8A + 2B + (n − 1)(−2A − B)
= 8A + 2B − 2An − Bn + 2A + B
= 10A + 3B − 2An − Bn
= (10 − 2n)A + B(3 − n)
Coefficient of A =  twice the coefficient of B
⇒ 10 − 2n = 2(3 − n)
⇒ 10 − 2n = 6 − 2n
⇒ 10 = 6
This is not possible.
⇒ no value of n exists for the given condition.
There is no value of n for which coefficient of A is twice the coefficient of B.

The 7th term of a geometric progression (GP) can be found using the formula:

Tn=a⋅rⁿ⁻¹

Where:

• Tn​ is the n-th term,
• a is the first term,
• r is the common ratio,
• n is the term number.

In the given GP: 2, 6, 18, ...

• The first term a=2
• The common ratio r=6/2=3

Now, substitute into the formula for the 7th term:

T₇ = 2⋅3⁷⁻¹ = 2⋅3⁶ = 2⋅729 = 1458

So, the 7th term is 1458

Let d be the common difference of the A.P.
Then a_2r = a_{2r − 1} + d

Given a_p = 1/q
a + (p − 1)d = 1/q
aq + (pq − q)d=1…(1)
Similarly, we get
ap + (pq − p)d = 1…(2)
From (1) and (2), we get
aq + (pq − q)d = ap + (pq − p)d
aq − ap = d[pq − p − pq + q]
a(q − p) = d(q − p)
∴ a = d
Equation (1) becomes,
dq + pqd − dq = 1

2 b = a + c, so the given expression is (a + a + c − c)(a + c + c − a)(2b − b) = 4abc