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The correct option is (c) 93

The explanation is: We know, Sn = a(R^N-1)/(r-1)

Here a = 3, r = 2 and n = 5

S5 = 3 (2^5-1) / (2-1) = 3(32 -1) = 3*31 = 93.

The correct answer is (b) -1

The EXPLANATION: an = an – 1–1, N > 2

=> a3 = a2 – 1 = 2 – 1 = 1

=> a4 = a3 – 1 = 1 – 1 = 0

=> a5 = a4 – 1 = 0 – 1 = -1.

Correct answer is (C) 17

The best explanation: SUM of the series 2+3+5+7 is finite because given series has finite NUMBER of terms. The sum of given 4 terms i.e. 17.

Correct choice is (c) 2,4,6,8,10

Explanation: Since SEQUENCE 2,4,6,8,10 CONTAINS limited number of terms so, it is finite sequence. Rest all are infinite sequences.

Right CHOICE is (b) 1185

For explanation I WOULD say: 6^2+7^2+………………..…..+15^2

= (1^2+2^2+3^2+……..+15^2) – (1^2+2^2+3^2+4^2+5^2)

= 15*16*31/6 – 5*6*11/6

= 1240-55 = 1185.

Right option is (c) 7

For EXPLANATION: In the given G.P., a=25 and r = 125/25 = 5.

Given, an = 390625 => ar^n-1 = 390625

=>25*5^n-1 = 390625

=> 5^n-1 = 390625/25 = 15625 = 5^6

=> n-1 = 6 => n=7.

Correct CHOICE is (b) 840

To explain: Given, a=20, a12= 120, n=12.

Sn = \(\frac{n}{2}\) (a+l) => S12 = \(\frac{12}{2} (20+120)\) = 6*140 = 840.

Correct option is (a) True

To EXPLAIN: Yes, 1,1,2,3,5, ………… is a Fibonacci SEQUENCE because it follows the recurrence relation

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

Right answer is (a) 9:1

The explanation is: (A.M.)/(G.M.) = 5/3

=> \(\frac{a+B}{2\sqrt{ab}} = \frac{5}{3}\)

Applying componendo and dividendo RULE, we get

=> \(\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}} = \frac{8}{2}\)

=> \((\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}})^2=4 \)

=> \((\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}})^1=2\)

=> \((\frac{\sqrt{a}}{\sqrt{b}})^1=3\)

Again applying componendo and dividendo rule, we get

a/b = (3/1)^2 = 9. So, a:b =9:1.

Correct option is (b) geometric PROGRESSION

For explanation I WOULD say: A SEQUENCE is called geometric progression if an+1 = an * r where a1 is the FIRST term and r is common ratio.

Correct option is (a) 63/32

To explain I would say: GIVEN SERIES is G.P. with first term 1 and common ratio 1/2.

We KNOW, Sn = a(1-rn)/(1-r) for r<1.

S6 = 1(1-(1/2)^6) / (1-1/2) = (1-1/64) / (1/2) = 63*2/64 = 63/32.

The correct OPTION is (b) 12 and 3

The explanation: We know, A.M. of two NUMBERS a and b is (a + b)/2

=> (a + b)/2 = 15/2 => a + b = 15.

Also, G.M. of two numbers a and b is \(\sqrt{ab}\)

=> \(\sqrt{ab}\) = 6 => ab = 36.

=> a(15-a) = 36 => a=3 or 12.

For a=3, b=12.

For a=12, b=3.

So, the two numbers are 3 and 12.

Right option is (d) 124

To explain: In the GIVEN G.P., a=4 and r=8/4=2.

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

=>S5 = 4(2^5-1) / (2-1) = 4*31 = 124.

The CORRECT choice is (a) 2556

Best explanation: 1^3+3^3+5^3+…………………………..+11^3

= (1^3+2^3+3^3+……+11^3) – (2^3+4^3+6^3+8^3+10^3)

= (1^3+2^3+3^3+……11^3) – 2^3(1^3+23+3^3+4^3+5^3)

= (11*12/2)^2 – 8(5*6/2)^2

= 66^2-8*15^2

= 4356 – 1800

= 2556.

The CORRECT OPTION is (b) A≥G

Easiest explanation: We know, A.M. of TWO numbers a and b is (a + b)/2

Also, G.M. of two numbers a and b is \(\sqrt{ab}\)

A-G = (a + b)/2 – \(\sqrt{ab} = ((a + b) – 2\sqrt{ab})/2 = (\sqrt{a} – \sqrt{b})^2\) / 2 ≥ 0

So, A≥G.

The CORRECT choice is (c) 9:31

To explain: Let a, a’ be the first terms and d, d’ be the COMMON differences of 2 A.P.’s respectively.

Given, \(\FRAC{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2a’+(n-1)d’]} = \frac{2n+3}{7n+5}\)

=>\(\frac{a+(n-1)d/2}{a’+(n-1) d’/2} = \frac{2n+3}{7n+5}\)

If we have to find ratio of 9^th terms then (n-1)/2 =8 => n=17

=>\(\frac{a+8d}{a’+8d’} = \frac{2*17+3}{7*17+5} = \frac{34+3}{119+5} = \frac{36}{124}\) = 9/31.

Correct OPTION is (d) 372

To EXPLAIN: We KNOW, a=20, d=2, an=42.

a+(n-1) d = 42 => 20 + 2(n-1) = 42

=>2(n-1) = 42-20=22 => n-1 = 11 => n=12.

Sn = \(\FRAC{n}{2}\) (a+l) => S12 = \(\frac{12}{2}\) (20+42) = 6*62 = 372.

The correct option is (C) 1408

To EXPLAIN: From the given A.P., a=1 and d=7-1 = 6.

We know, Sn = \(\frac{n}{2} (2a+(n-1)d)\)

S22 = \(\frac{22}{2} (2*1+(22-1)6)\) = 11(2+21*6) = 11(2+126) = 11*128 = 1408.

The correct choice is (a) an = a + (n-1) d

To explain I WOULD SAY: Since every term of an A.P. is incremented by common DIFFERENCE d.

i.e. an+1 = an + d = an-1 + 2d = ……. = A1 + n*d

or an = a + (n-1) d

Correct ANSWER is (d) 128

The best I can explain: Let G.P. be 4, G1, G2, G3, 512.

=>a=4 and A5 = a*R^4 = 512 => 4*r^4 = 512 => r^4 = 512/4 = 128 => r = 4.

G1 = a2 = a * r = 4*4 = 16.

G2 = G1 * r = 16 * 4 = 64.

G3 = G2 * r = 64*4 = 256.

Right OPTION is (a) 0

The explanation: a=171 and d=162-171 = -9.

an<=0

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

=>180-9n <=0

=>9n >= 180

=> n >= 20 => n=20 for first non-positive TERM.

First negative term is 171+(19) (-9) = 171-171 = 0.

The correct ANSWER is (b) 3

The EXPLANATION: Given, an = 3n.

We know, d = an-an-1 = 3n – 3(n-1) = 3.

Correct OPTION is (B) False

Easiest explanation: Sequence and progression are DIFFERENT things. When sequence follow a specified pattern, it is said to be a progression.

Correct answer is (B) \(\frac{n(n+1)(2n-5)}{6}\)

To explain: GIVEN, n^th TERM is n(n-2)

So, ak = k(k-2)

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

\(\sum_{i=0}^n a_k = \sum_{i=0}^n k^2 – 2 \sum_{i=0}^n k = \frac{n(n+1)(2n+1)}{6} – 2\frac{n(n+1)}{2} = \frac{n(n+1)(2n-5)}{6}\).

Right OPTION is (c) \(\frac{n(n+1)(2n+1)}{6}\)

Explanation: Sum of SQUARES of first n terms = 1^2+2^2+3^2+……………+n^2

k^3–(k – 1)^3=3k^2–3k + 1

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

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

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

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

Right CHOICE is (b) \(\frac{8±3\sqrt{7}}{1}\)

BEST explanation: We know, G.M. of two numbers a and b is √ab.

So, a + b = 4 √ab

Squaring we GET, a^2+b^2 = 16ab

=>(a/b) + (b/a) = 16

Let x = a/b.

So, x + 1/x = 16 => x^2 – 16x + 1 = 0

=>x = \(\frac{16±\sqrt{256-4}}{2} = \frac{16±\sqrt{252}}{2} = \frac{16±6\sqrt{7}}{2} = \frac{8±3\sqrt{7}}{1}\).

Correct option is (b) 2/3

To explain: LET THREE terms be a/R, a, a*r.

Product = 27 => (a/r) (a) (a*r) = 27 => a^3 = 27

=>a = 3.

Sum = 21/2 => (a / r + a + a*r) = 21/2 => a (1 / r + 1 + 1*r) = 21/2

=> (1 / r + 1 + 1*r) = (21/2)/3 = 7/2

=> (r^2 + r + 1) = (7/2) r => r^2 – (5/2) r +1 = 0

=> r = 2 and 1/2.

Terms are 3/2, 3, 3*2 i.e. 3/2, 3, 6.

Right answer is (c) 60

Best explanation: GIVEN, a3 = 6 and A5 = 12.

=> a + 2d = 6 and a + 4d = 12

=> 2d = 6 => d=3.

=> a + 2*3 = 6 => a=0

So, A21 = a + 20 d = 0 + 20*3 = 60.

The correct answer is (a) True

Explanation: LET X be the constant which is added 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 = a’ + (n-1) d which is n^th term of an A.P. If x is negative it is CASE of subtraction.

The correct option is (d) 30

To EXPLAIN I would SAY: Since 2, 4 and 6 all are EVEN numbers so, given series involve all even number terms.

The next TWO terms will be 8 and 10 so, sum will be 2+4+6+8+10 = 30.

The CORRECT answer is (d) 66

For explanation: an = 4n+6 and n=15

=>A15 = 4*15 + 6 = 60+6 = 66.

Correct OPTION is (d) 966

Explanation: GENERAL term of above series is ak = 2K*(k^2+2) = 2k^3+4k

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

\(\sum_{i=0}^na_k = 2\sum_{i=0}^nk^3 + 4\sum_{i=0}^nk = 2(\FRAC{n(n+1)}{2})^2 + 4\frac{n(n+1)}{2}\)

= n^2(n+1)^2/2+2n(n+1)

= 36*49/2 + 2*6*7

= 966.

The correct option is (b) 6

Explanation: We KNOW, geometric MEAN of two numbers a and b is GIVEN by

G.M. = \(\sqrt{a*b}\)

So, G.M. of 3 and 12 is \(\sqrt{3*12} = \sqrt{36}\) = 6.

Right OPTION is (a) ARITHMETIC progression

Easy EXPLANATION: A sequence is called arithmetic progression if an+1 = an + d where a1 is the first TERM and d is common difference.

Correct OPTION is (d) 11

For EXPLANATION I would say: Since 2,3,5 and 7 all are consecutive PRIME NUMBERS so, it is a sequence of prime numbers. Prime NUMBER next to 7 is 11. So, 2,3,5,7,11.

Correct option is (c) 385

The best I can explain: We KNOW, sum of SQUARES of FIRST n terms is GIVEN by \(\frac{(n(n+1)(2n+1))}{6}\).

Here, n=10. So, sum = 10*11*21/6 = 385.