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51.

How many terms of G.P. 2,4,8,16, …………… are required to give sum 254?(a) 4(b) 5(c) 6(d) 7This question was posed to me in an international level competition.This key question is from Geometric Progression(G.P.) in chapter Sequences and Series of Mathematics – Class 11

Answer» RIGHT answer is (d) 7

To EXPLAIN I would say: a=2 and r = 4/2 = 2.

We KNOW, SN = a(r^n-1)/(r-1)

2(2^n-1) / (2-1) = 254

=>2^n-1 = 127 => 2^n = 128 = 2^7

=> n=7.
Discussion
52.

If a sequence is in the form 2*5^n then which of the following may be the sequence?(a) Arithmetic progression(b) Geometric Progression(c) Harmonic Progression(d) Special ProgressionI have been asked this question in examination.The above asked question is from Geometric Progression(G.P.) topic in portion Sequences and Series of Mathematics – Class 11

Answer»

Right answer is (b) GEOMETRIC Progression

To EXPLAIN I WOULD say: If an = 2*5^n then

a1 =10, a2 = 50, a3=250.

This is a geometric progression with first TERM 10 and common ratio 5.
Discussion
53.

In A.P. 171, 162, 153, ………. Find first negative term.(a) 0(b) -2(c) -6(d) -9I got this question by my school teacher while I was bunking the class.The query is from Arithmetic Progression(A.P.) topic in portion Sequences and Series of Mathematics – Class 11

Answer»

Right option is (d) -9

For EXPLANATION I would say: a=171 and d=162-171 = -9.

an<0

=>171+(N-1) (-9) < 0

=>180-9n < 0

=>9n > 180

=>n > 20 => n=21 for first negative term.

First negative term is 171+(20) (-9) = 171-180 = -9
Discussion
54.

A series can also be denoted by symbol ____________(a) π an(b) ∑an(c) Φ an(d) θ anThis question was addressed to me in semester exam.This intriguing question originated from Series topic in division Sequences and Series of Mathematics – Class 11

Answer»

The correct option is (b) ∑an

To explain: When we use addition between the terms of SEQUENCE, it is said to be series.

We know that addition can ALSO be WRITTEN in the form of SIGMA so, series can also be DENOTED by ∑an.
Discussion
55.

Find the sum 1^3+2^3+3^3+……………+8^3.(a) 1225(b) 1184(c) 1475(d) 1296I had been asked this question in final exam.My question is from Sum to n Terms of Special Series topic in portion Sequences and Series of Mathematics – Class 11

Answer»

The CORRECT choice is (d) 1296

To EXPLAIN: We know, sum of cubes of first N terms is given by \((\frac{n(n+1)}{2})^2\).

Here, n=8. So, sum = (8*9/2)^2 = 1296.
Discussion
56.

Which of the following relation gives Fibonacci sequence?(a) an = an-1 + an-2(b) an-1 = an + an-2(c) an-2 = an + an-1(d) an = an+1 + an-2The question was posed to me by my school principal while I was bunking the class.Enquiry is from Sequences topic in section Sequences and Series of Mathematics – Class 11

Answer»

The CORRECT answer is (a) an = an-1 + an-2

Explanation: an = an-1 + an-2, n>2.

This is a recurrence relation which GIVES FIBONACCI sequence.
Discussion
57.

Find the sum of series up to 6^th term whose n^th term is given by n^2 + 3^n.(a) 91(b) 1284(c) 1183(d) 1092I had been asked this question in exam.I'd like to ask this question from Sum to n Terms of Special Series topic in division Sequences and Series of Mathematics – Class 11

Answer»

Right choice is (C) 1183

To explain: Given, n^th TERM is n^2 + 3^n

So, AK = k^2 + 3^k

Taking SUMMATION from k=1 to k=n on both sides, we get

\(\sum_{i=0}^na_k = \sum_{i=0}^nk^2 + \sum_{i=0}^n3^k\)

\(\sum_{i=0}^nk^2 = n(n+1) (2n+1)/6\)

\(\sum_{i=0}^n3^k = 3*(3^n-1)/ (3-1) = (3/2) (3^n-1)\)

So, \(\sum_{i=0}^na_k = n(n+1) (2n+1)/6 + (3/2) (3^n-1)\)

SUM up to 6^th term = 6*7*13/6 + (3/2) (3^6-1) = 91+1092 = 1183.
Discussion
58.

Complete 2, 4, 6, 8, _____________(a) 10(b) 9(c) 13(d) 11This question was addressed to me during an interview for a job.Origin of the question is Sequences in section Sequences and Series of Mathematics – Class 11

Answer» CORRECT answer is (a) 10

Explanation: SINCE 2,4,6 and 8 are even numbers so it is a sequence of even numbers. Even number NEXT to 8 is 10. So, 2,4,6,8,10.
Discussion
59.

If an A.P. is 3,5,7,9……. Find the 12^th term of the A.P.(a) 12(b) 21(c) 22(d) 25I got this question by my college director while I was bunking the class.The query is from Arithmetic Progression(A.P.) in division Sequences and Series of Mathematics – Class 11

Answer» CORRECT ANSWER is (d) 25

To elaborate: From the given A.P., a=3 and d=5-3 =2.

We know, an= a + (n-1) d => A12 = a+11d = 3+11*2 = 3+22 = 25.
Discussion
60.

Find the sum of cubes of first n terms.(a) \(\frac{n(n+1)}{2}\)(b) \((\frac{n(n+1)}{2})^3\)(c) \(\frac{n(n+1)(2n+1)}{6}\)(d) \((\frac{n(n+1)}{2})^2\)I got this question in an online quiz.Origin of the question is Sum to n Terms of Special Series in section Sequences and Series of Mathematics – Class 11

Answer»

Right answer is (C) \(\FRAC{N(n+1)(2n+1)}{6}\)

Best explanation: Sum of CUBES of first n terms = 1^3+2^3+3^3+……………+n^3

(k + 1)^4–k^4 = 4k^3 + 6k^2 + 4k + 1.

On substituting k = 1, 2, 3, ……, n and adding we get,

4n^3+n^4+6n^2+4n = \(4\sum_{i=0}^n k^3 + 6 \sum_{i=0}^n k^2 + 4 \sum_{i=0}^n k + n\)

4n^3+n^4+6n^2+4n = \(4\sum_{i=0}^n k^3 + 6\frac{(n(n+1)(2n+1))}{6} + 4\frac{n(n+1)}{2} + n\)

\(\sum_{i=0}^n k^3 = (\frac{n(n+1)}{2})^2\).
Discussion
61.

Find the sum up to 7^th term of series 2+3+5+8+12+………………….(a) 70(b) 490(c) 340(d) 420The question was asked in an interview for job.This interesting question is from Sum to n Terms of Special Series in portion Sequences and Series of Mathematics – Class 11

Answer»

The correct choice is (a) 70

Easiest EXPLANATION: Sn = 2+3+5+8+12+……………………………+ an

Sn = 2+3+5+8+12+ ……. + an-1 + an

Subtracting we get, 0 = 2+1+2+3+4+………………………….. – an

=>an = 2+1+2+3+4+…………….+(N-1) = 2+(n-1)n/2 = (1/2) (n^2-n+4)

n^th TERM is (1/2) (n^2-n+4)

So, AK = (1/2) (k^2-k+4)

Taking summation from k=1 to k=n on both sides, we get

\(\sum_{i=0}^na_k = (1/2)\sum_{i=0}^nk^2 – (1/2)\sum_{i=0}^nk + 2n\) = n(n+1) (2n+1)/(2*6) – n(n+1)/4 + 2n

Here, n=7. So, \(\sum_{i=0}^na_k\) = (7*8*15)/12 – (7*8)/4 + 2*7 = 70.
Discussion
62.

If sum of n terms of an A.P. is n^2+5n then find general term.(a) n+1(b) 2n(c) 3n(d) n^2+3nI had been asked this question in an internship interview.My enquiry is from Arithmetic Progression(A.P.) topic in chapter Sequences and Series of Mathematics – Class 11

Answer»

The correct option is (B) 2N

The explanation is: GIVEN, Sn = n^2+5n

We KNOW, an = Sn– Sn-1 = (n^2+5n) – ((n-1)^2+5(n-1)) = (n^2+5n) – (n^2+1-2n+5n-1) = 2n.
Discussion
63.

What is the third term of Fibonacci sequence?(a) 0(b) 1(c) 2(d) 3This question was posed to me in an interview.I'd like to ask this question from Sequences in section Sequences and Series of Mathematics – Class 11

Answer»

The correct option is (C) 2

To explain I WOULD say: a1=1 and a2=1.

an = an-1 + an-2, n>2.

This is a recurrence RELATION which gives FIBONACCI sequence.

=>a3=a1+a2=1+1=2.
Discussion
64.

If a constant is multiplied to A.P. then resulting sequence is also an A.P.(a) True(b) FalseI had been asked this question during a job interview.My doubt is from Arithmetic Progression(A.P.) topic in chapter Sequences and Series of Mathematics – Class 11

Answer»

The correct answer is (a) True

To explain I WOULD say: Let X be the CONSTANT which is MULTIPLIED to each term of an A.P., then an’ = an * x and a’ = a * x. So n^th term will be an’ = an * x = (a+(n-1) d) * x = (a * x) + (n-1) d x = a’ + (n-1) x d which is n^th term of an A.P.
Discussion
65.

Insert 4 numbers between 2 and 22 such that the resulting sequence is an A.P.(a) 4, 8, 12, 16(b) 5, 9, 13, 17(c) 4, 10, 15, 19(d) 6, 10, 14, 18I have been asked this question in a job interview.This is a very interesting question from Arithmetic Progression(A.P.) topic in division Sequences and Series of Mathematics – Class 11

Answer»

The correct CHOICE is (d) 6, 10, 14, 18

Easy explanation: Let A.P. be 2, A1, A2, A3, A4, 22.

=>a=2 and a6 = a+5d = 22 => 2+5*d=22 => d=4.

A1 = a2 = a + d = 2 + 4 = 6.

A2 = A1 + d = 6 + 4 = 10.

A3 = 10 + 4 = 14.

A4 = 14 + 4 = 18.
Discussion
66.

\(\sum_{i=1}^4\) 2n+3 = _____________________(a) 5(b) 12(c) 21(d) 32This question was posed to me by my college professor while I was bunking the class.Query is from Series in portion Sequences and Series of Mathematics – Class 11

Answer»

Correct answer is (d) 32

The best EXPLANATION: a1 = 2*1+3 = 5, a2 = 2*2+3 = 7, A3 = 2*3+3 = 9, a4 = 2*4+3 = 11.

Sum = 5+7+9+11 = 32.
Discussion