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JEE Advanced Level Test: Sequences And Series- 1 - JEE MCQ


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22 Questions MCQ Test - JEE Advanced Level Test: Sequences And Series- 1

JEE Advanced Level Test: Sequences And Series- 1 for JEE 2024 is part of JEE preparation. The JEE Advanced Level Test: Sequences And Series- 1 questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Level Test: Sequences And Series- 1 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Level Test: Sequences And Series- 1 below.
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JEE Advanced Level Test: Sequences And Series- 1 - Question 1

The first term of an A.P. of consecutive integer is p2 + 1. The sum of (2p + 1) terms of this series can be expressed as

Detailed Solution for JEE Advanced Level Test: Sequences And Series- 1 - Question 1

a = p2 + 1
Since series is of consecutive integer
Sum of (2p+1) terms = n/2 (2a+(n−1)d)
= [(2p+1)/2][2(p2 + 1)+(2p + 1 − 1)1]
= [(2p + 1)/2][2p2 + 2 + 2p]
=(2p + 1)(p2 + p + 1)
= 2p3 + 2p2 + 2p + p2 + p + 1
= p3 +p3 + 1 + 3p2 +3p
= p3 + (p+1)3

JEE Advanced Level Test: Sequences And Series- 1 - Question 2

If a1, a2, a3,........ are in A.P. such that a1 + a5 + a10 + a15 + a20 + a24 = 225, then a1 + a2 + a3 + ......+ a23 + a24 is equal to

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JEE Advanced Level Test: Sequences And Series- 1 - Question 3

If x ∈ R, the numbers 51+x + 51-x, a/2, 25x + 25-x form an A.P. then `a' must lie in the interval;

Detailed Solution for JEE Advanced Level Test: Sequences And Series- 1 - Question 3


JEE Advanced Level Test: Sequences And Series- 1 - Question 4

The third term of a G.P. is 4. The product of the first five terms is

Detailed Solution for JEE Advanced Level Test: Sequences And Series- 1 - Question 4

Let a and r the first term and common ratio, respectively.

JEE Advanced Level Test: Sequences And Series- 1 - Question 5

If S is the sum of infinity of a G.P. whose first term is `a', then the sum of the first n terms is

Detailed Solution for JEE Advanced Level Test: Sequences And Series- 1 - Question 5

Given that first term =a
S= sum to infinity of a G.P= a/(1−r)
r=1−a/S
Sn=a(1−rn)/(1-r)
Sn=S(1−rn)
Sn=S[1−(1−a/S)n]

JEE Advanced Level Test: Sequences And Series- 1 - Question 6

The sum of the series  +  +  + ....... +  is

Detailed Solution for JEE Advanced Level Test: Sequences And Series- 1 - Question 6

JEE Advanced Level Test: Sequences And Series- 1 - Question 7

For a sequence {an}, a= 2 and . Then  is

Detailed Solution for JEE Advanced Level Test: Sequences And Series- 1 - Question 7

JEE Advanced Level Test: Sequences And Series- 1 - Question 8

a, b be the roots of the equation x2 – 3x + a = 0 and g, d the roots of x2 – 12x + b = 0 and numbers a, b, g, d (in this order) form an increasing G.P., then

Detailed Solution for JEE Advanced Level Test: Sequences And Series- 1 - Question 8

JEE Advanced Level Test: Sequences And Series- 1 - Question 9

If 3 + 1/4 (3 + d) + 1/42 (3 + 2d) + .... + upto ∞ = 8, then the value of d is

Detailed Solution for JEE Advanced Level Test: Sequences And Series- 1 - Question 9

S=(3 + 3/4 + 3/42 +....)+(d/4 + 2d/42 + 3d/42 + ....)
S = 3(1 + 1/4 + 1/42 + ..)+(s) (lets's say)
s = d/4 + 2d/42 + 3d/43 + ..(1)
s/4 = d/42 + 2d/43 + ...(2)
subtrating  (1) - (2)
3s/4 = d/4 + d/42 + d/43 + ....
s = d/3(1 + 1/4 + 1/42 + ..)
s= 4d/9
S=3(3/4) + 4d/9
S = 4 + 4d/9 = 8
d = 9

JEE Advanced Level Test: Sequences And Series- 1 - Question 10

The sum  is equal to

Detailed Solution for JEE Advanced Level Test: Sequences And Series- 1 - Question 10

JEE Advanced Level Test: Sequences And Series- 1 - Question 11

In a G.P. of positive terms, any term is equal to the sum of the next two terms. The common ratio of the G.P. is

Detailed Solution for JEE Advanced Level Test: Sequences And Series- 1 - Question 11

JEE Advanced Level Test: Sequences And Series- 1 - Question 12

If  + ...... up to ∞ =, then  + ..... =

Detailed Solution for JEE Advanced Level Test: Sequences And Series- 1 - Question 12

(1/12 + 1/22 + 1/32 + ........∞) = π2/6
⇒ (1/12 + 1/32 + 1/52 + ........∞) + (1/22 + 1/42 + 1/62 + ........∞) = π2/6
=> (1/12 + 1/32 + 1/52 + ........∞) + 1/22(1/12 + 1/22 + 1/32 + ........∞) = π2/6
=> (1/12 + 1/32 + 1/52 + ........∞) + 1/222/6) = π2/6
=> (1/12 + 1/32 + 1/52 + ........∞) + π2/24 = π2/6
= π2/8

JEE Advanced Level Test: Sequences And Series- 1 - Question 13

Sum of the series S = 12 – 22 + 32 – 42 + ..... – 20022 + 20032 is

Detailed Solution for JEE Advanced Level Test: Sequences And Series- 1 - Question 13

S=12−22+32−42+......−20022+20032

JEE Advanced Level Test: Sequences And Series- 1 - Question 14

If x =, y =, z =where a, b, c are in AP and |a| < 1, |b| < 1, |c| < 1, then x, y, z are in

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JEE Advanced Level Test: Sequences And Series- 1 - Question 15

The sum to n term of the series 1(1!) + 2(2!) + 3(3!) + ....

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JEE Advanced Level Test: Sequences And Series- 1 - Question 16

The sum to 10 terms of the series is

Detailed Solution for JEE Advanced Level Test: Sequences And Series- 1 - Question 16

JEE Advanced Level Test: Sequences And Series- 1 - Question 17

If p is positive, then the sum to infinity of the series,  is

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JEE Advanced Level Test: Sequences And Series- 1 - Question 18

The positive integer n for which 2 × 22 + 3 × 23 + 4 × 24 + ..... + n × 2n = 2n + 10 is

Detailed Solution for JEE Advanced Level Test: Sequences And Series- 1 - Question 18

JEE Advanced Level Test: Sequences And Series- 1 - Question 19

If x > 0, and log2 x + log2  + log2  + log2  + log2  + ....= 4, then x equals

JEE Advanced Level Test: Sequences And Series- 1 - Question 20

If a, b, c are in A.P. p, q, r are in H.P. and ap, bq, cr are in G.P., then  is equal to

Detailed Solution for JEE Advanced Level Test: Sequences And Series- 1 - Question 20

JEE Advanced Level Test: Sequences And Series- 1 - Question 21

12 + 22 + ......+ n2 = 1015, then value of n is

JEE Advanced Level Test: Sequences And Series- 1 - Question 22

If a and b are pth and qth terms of an AP, then the sum of its (p + q) terms is

Detailed Solution for JEE Advanced Level Test: Sequences And Series- 1 - Question 22


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