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Sequences and Series

JEE Mathematics · 201 questions · Page 19 of 21 · Click an option or "Show Solution" to reveal answer

Q181
Let SnS_n denote the sum of the first nn terms of an arithmetic progression. If S10=390S_{10}=390 and the ratio of the tenth and the fifth terms is 15:715: 7, then S15−S5\mathrm{S}_{15}-\mathrm{S}_5 is equal to :
A 800
B 890
C 790
D 690
Correct Answer
Option C
Solution

To solve this problem, we will start by using the properties of an arithmetic progression (AP).

The sum of the first nn terms of an AP can be calculated using the formula:

Sn=n2(2a+(n−1)d)S_n = \frac{n}{2} (2a + (n-1)d)

where SnS_n is the sum of the first nn terms, aa is the first term, and dd is the common difference between the terms.

Given the information:

S10=390S_{10} = 390

We can plug n=10n=10 into the sum formula to get:

S10=102(2a+(10−1)d)S_{10} = \frac{10}{2} (2a + (10-1)d)
390=5(2a+9d)390 = 5(2a + 9d)
390=10a+45d390 = 10a + 45d
78=2a+9d.........(1)78 = 2a + 9d \quad .........\text{(1)}

Next, we're given the ratio of the tenth term (T10T_{10}) to the fifth term (T5T_5):

T10T5=157\frac{T_{10}}{T_5} = \frac{15}{7}

The nnth term of an AP is given by:

Tn=a+(n−1)dT_n = a + (n-1)d

So, for the tenth term:

T10=a+(10−1)d=a+9dT_{10} = a + (10-1)d = a + 9d

And for the fifth term:

T5=a+(5−1)d=a+4dT_5 = a + (5-1)d = a + 4d

Now we can write the ratio as:

a+9da+4d=157\frac{a + 9d}{a + 4d} = \frac{15}{7}
7(a+9d)=15(a+4d)7(a + 9d) = 15(a + 4d)
7a+63d=15a+60d7a + 63d = 15a + 60d
63d−60d=15a−7a63d - 60d = 15a - 7a
3d=8a.........(2)3d = 8a \quad .........\text{(2)}

Now we have two equations (1) and (2):

78=2a+9d(1)78 = 2a + 9d \quad \text{(1)}
3d=8a(2)3d = 8a \quad \text{(2)}

We can solve these equations simultaneously. From equation (2):

d=83ad = \frac{8}{3}a

Plugging this back into (1):

78=2a+9(83a)78 = 2a + 9\left(\frac{8}{3}a\right)
78=2a+24a78 = 2a + 24a
78=26a78 = 26a
a=3a = 3

Now we can find dd:

d=83ad = \frac{8}{3}a
d=83×3d = \frac{8}{3} \times 3
d=8d = 8

Now we can find S15S_{15} and S5S_5 using the formula for the sum of an AP. For S15S_{15}:

S15=152(2⋅3+(15−1)⋅8)S_{15} = \frac{15}{2} (2 \cdot 3 + (15-1) \cdot 8)
S15=152(6+14⋅8)S_{15} = \frac{15}{2} (6 + 14 \cdot 8)
S15=152(6+112)S_{15} = \frac{15}{2} (6 + 112)
S15=152⋅118S_{15} = \frac{15}{2} \cdot 118
S15=15⋅59S_{15} = 15 \cdot 59
S15=885S_{15} = 885

For S5S_5:

S5=52(2⋅3+(5−1)⋅8)S_5 = \frac{5}{2} (2 \cdot 3 + (5-1) \cdot 8)
S5=52(6+4⋅8)S_5 = \frac{5}{2} (6 + 4 \cdot 8)
S5=52(6+32)S_5 = \frac{5}{2} (6 + 32)
S5=52⋅38S_5 = \frac{5}{2} \cdot 38
S5=5⋅19S_5 = 5 \cdot 19
S5=95S_5 = 95

The difference S15−S5S_{15} - S_{5} is:

S15−S5=885−95S_{15} - S_{5} = 885 - 95
S15−S5=790S_{15} - S_{5} = 790

Therefore, the correct answer is Option C, which is 790.

Q182
If log⁡ea,log⁡e b,log⁡ec\log _e \mathrm{a}, \log _e \mathrm{~b}, \log _e \mathrm{c} are in an A.P. and log⁡ea−log⁡e2 b,log⁡e2 b−log⁡e3c,log⁡e3c−log⁡e\log _e \mathrm{a}-\log _e 2 \mathrm{~b}, \log _e 2 \mathrm{~b}-\log _e 3 \mathrm{c}, \log _e 3 \mathrm{c} -\log _e a are also in an A.P, then a:b:ca: b: c is equal to
A 6:3:26: 3: 2
B 9:6:49: 6: 4
C 25:10:425: 10: 4
D 16:4:116: 4: 1
Correct Answer
Option B
Solution
log⁡ea,log⁡eb,log⁡ec\log _{\mathrm{e}} \mathrm{a}, \log _{\mathrm{e}} \mathrm{b}, \log _{\mathrm{e}} \mathrm{c}

are in A.P.

∴b2=ac\therefore \mathrm{b}^2=\mathrm{ac}

..... (i) Also

log⁡e(a2b),log⁡e(2b3c),log⁡e(3ca) are in A.P. (2b3c)2=a2b×3cabc=32\begin{aligned} & \log _e\left(\frac{a}{2 b}\right), \log _e\left(\frac{2 b}{3 c}\right), \log _e\left(\frac{3 c}{a}\right) \text{ are in A.P. } \\ & \left(\frac{2 b}{3 c}\right)^2=\frac{a}{2 b} \times \frac{3 c}{a} \\ & \frac{b}{c}=\frac{3}{2} \end{aligned}

Putting in eq. (i)

b2=a×2b3b^2=a \times \frac{2 b}{3}
ab=32a:b:c=9:6:4\begin{aligned} & \frac{\mathrm{a}}{\mathrm{b}}=\frac{3}{2} \\ & \mathrm{a}: \mathrm{b}: \mathrm{c}=9: 6: 4 \end{aligned}
Q183
The number of common terms in the progressions 4,9,14,19,……4,9,14,19, \ldots \ldots, up to 25th 25^{\text{th }} term and 3,6,9,12,……3,6,9,12, \ldots \ldots, up to 37th 37^{\text{th }} term is :
A 9
B 8
C 5
D 7
Correct Answer
Option D
Solution
4,9,14,19,…4,9,14,19, \ldots

, up to

25th 25^{\text{th }}

term

T25=4+(25−1)5=4+120=124\mathrm{T}_{25}=4+(25-1) 5=4+120=124
3,6,9,12,…3,6,9,12, \ldots

, up to

37th 37^{\text{th }}

term

T37=3+(37−1)3=3+108=111\mathrm{T}_{37}=3+(37-1) 3=3+108=111

Common difference of

Ist \mathrm{I}^{\text{st }}

series

d1=5\mathrm{d}_1=5

Common difference of

IInd \mathrm{II}^{\text{nd }}

series

d2=3\mathrm{d}_2=3

First common term

=9=9

, and their common difference

=15(LCM⁡=15\left(\operatorname{LCM}\right.

of

d1\mathrm{d}_1

and

d2)\left.\mathrm{d}_2\right)

then common terms are

9,24,39,54,69,84,999,24,39,54,69,84,99
Q184
 The 20th  term from the end of the progression 20,1914,1812,1734,…,−12914 is : \text{ The } 20^{\text{th }} \text{ term from the end of the progression } 20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots,-129 \frac{1}{4} \text{ is : }
A −115-115
B −100-100
C −110-110
D −118-118
Correct Answer
Option A
Solution
20,1914,1812,1734,……,−1291420,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots \ldots,-129 \frac{1}{4}

This is A.P. with common difference

d1=−1+14=−34−12914,…………...1914,20\begin{aligned} & d_1=-1+\frac{1}{4}=-\frac{3}{4} \\ & -129 \frac{1}{4}, \ldots \ldots \ldots \ldots . . .19 \frac{1}{4}, 20 \end{aligned}

This is also A.P.

a=−12914\mathrm{a}=-129 \frac{1}{4}

and

d=34\mathrm{d}=\frac{3}{4}

Required term ==

−12914+(20−1)(34)=−129−14+15−34=−115\begin{aligned} & -129 \frac{1}{4}+(20-1)\left(\frac{3}{4}\right) \\ & =-129-\frac{1}{4}+15-\frac{3}{4}=-115 \end{aligned}
Q185
Let a,ar,ar2a, a r, a r^2, ............ be an infinite G.P. If ∑n=0∞arn=57\sum\limits_{n=0}^{\infty} a r^n=57 and ∑n=0∞a3r3n=9747\sum\limits_{n=0}^{\infty} a^3 r^{3 n}=9747, then a+18ra+18 r is equal to
A 27
B 38
C 31
D 46
Correct Answer
Option C
Solution
∑n=0∞arn=57⇒a1−r=57.... (i)∑n=0∞a3r3n=9747⇒a31−r3=9747.... (ii)\begin{array}{ll} \sum_{n=0}^{\infty} a r^n=57 & \Rightarrow \frac{a}{1-r}=57 \quad \text{.... (i)}\\ \sum_{n=0}^{\infty} a^3 r^{3 n}=9747 & \Rightarrow \frac{a^3}{1-r^3}=9747 \quad \text{.... (ii)} \end{array}
(1−r3)(1−r)3=(57)39747=19⇒(1−r)(1+r+r2)(1−r)3=19⇒18r2−39r+18=0⇒r=23,32 (rejected) ∴a=19a+18r=19+12=31\begin{aligned} & \frac{\left(1-r^3\right)}{(1-r)^3}=\frac{(57)^3}{9747}=19 \\ & \Rightarrow \quad \frac{(1-r)\left(1+r+r^2\right)}{(1-r)^3}=19 \\ & \Rightarrow \quad 18 r^2-39 r+18=0 \\ & \Rightarrow \quad r=\frac{2}{3}, \frac{3}{2} \text{ (rejected) } \\ & \therefore \quad a=19 \\ & \quad a+18 r \\ & \quad=19+12=31 \end{aligned}
Q186
In an increasing geometric progression of positive terms, the sum of the second and sixth terms is 703\dfrac{70}{3} and the product of the third and fifth terms is 49. Then the sum of the 4th ,6th 4^{\text{th }}, 6^{\text{th }} and 8th 8^{\text{th }} terms is equal to:
A 78
B 96
C 91
D 84
Correct Answer
Option C
Solution

Let's denote the first term of the geometric progression by

aa

and the common ratio by

rr

. The terms of the geometric progression can be written as follows: First term:

aa

Second term:

arar

Third term:

(ar2)(ar^2)

Fourth term:

(ar3)(ar^3)

Fifth term:

(ar4)(ar^4)

Sixth term:

(ar5)(ar^5)

Eighth term:

(ar7)(ar^7)

We are given two key pieces of information: 1. The sum of the second and sixth terms is

703\frac{70}{3}

:

ar+ar5=703ar + ar^5 = \frac{70}{3}

2. The product of the third and fifth terms is 49:

(ar2)⋅(ar4)=49(ar^2) \cdot (ar^4) = 49
a2r6=49a^2 r^6 = 49
a2=49r6a^2 = \frac{49}{r^6}
a=7r3a = \frac{7}{r^3}

Substituting

a=7r3a = \frac{7}{r^3}

into the first equation:

7r3⋅r+7r3⋅r5=703\frac{7}{r^3} \cdot r + \frac{7}{r^3} \cdot r^5 = \frac{70}{3}
7rr3+7r5r3=703\frac{7r}{r^3} + \frac{7r^5}{r^3} = \frac{70}{3}
7r2+7r21=703\frac{7}{r^2} + \frac{7r^2}{1} = \frac{70}{3}

Let

x=r2x = r^2

. Then:

7x+7x=703\frac{7}{x} + 7x = \frac{70}{3}

Multiply through by 3x to clear the denominator:

21+21x2=70x21 + 21x^2 = 70x

Rearrange into a standard quadratic equation:

21x2−70x+21=021x^2 - 70x + 21 = 0

Divide by 7 to simplify:

3x2−10x+3=03x^2 - 10x + 3 = 0

Solve this quadratic equation using the quadratic formula:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Where

a=3a = 3

,

b=−10b = -10

, and

c=3c = 3

. Thus:

x=10±100−366x = \frac{10 \pm \sqrt{100 - 36}}{6}
x=10±646x = \frac{10 \pm \sqrt{64}}{6}
x=10±86x = \frac{10 \pm 8}{6}
x=3x = 3

or

x=13x = \frac{1}{3}

Since

x=r2x = r^2

and

rr

is positive, we get

r=3r = \sqrt{3}

or

r=13r = \frac{1}{\sqrt{3}}

. We need to choose the value that results in positive, increasing terms: If

r=3r = \sqrt{3}

:

a=7r3=7(3)3=733=73⋅13=739a = \frac{7}{r^3} = \frac{7}{(\sqrt{3})^3} = \frac{7}{3\sqrt{3}} = \frac{7}{3} \cdot \frac{1}{\sqrt{3}} = \frac{7\sqrt{3}}{9}

Now we can determine the sum of the 4th, 6th, and 8th terms: The 4th term is:

ar3=739⋅33=7ar^3 = \frac{7\sqrt{3}}{9} \cdot 3\sqrt{3} = 7

The 6th term is:

ar5=739⋅93=21ar^5 = \frac{7\sqrt{3}}{9} \cdot 9\sqrt{3} = 21

The 8th term is:

ar7=739⋅27(3)=49ar^7 = \frac{7\sqrt{3}}{9} \cdot 27(\sqrt{3}) = 49

Adding these together:

(4th+6th+8thterms)=7+21+63=91(4th + 6th + 8th terms) = 7 + 21 + 63 = 91

Therefore, the sum of the 4th, 6th, and 8th terms is 91. Correct answer: Option C: 91

Q187
Let a1,a2,a3,…a_1, a_2, a_3, \ldots be in an A.P. such that ∑k=112a2k−1=−725a1,a1≠0\sum\limits_{k=1}^{12} a_{2 k-1}=-\dfrac{72}{5} a_1, a_1 \neq 0. If ∑k=1nak=0\sum\limits_{k=1}^n a_k=0, then nn is :
A 18
B 17
C 11
D 10
Correct Answer
Option C
Solution

Given: ∑k=112a2k−1=−725a1 \sum\limits_{k=1}^{12} a_{2k-1} = -\dfrac{72}{5} a_1 This means: a1+a3+⋯+a23=−725a1 a_1 + a_3 + \cdots + a_{23} = -\dfrac{72}{5} a_1 Express each term in A.P.: The general term of the A.P. is given by ak=a+(k−1)d a_k = a + (k-1)d .

Thus, for the odd indices: a1+(a+2d)+(a+4d)+⋯+(a+22d) a_1 + (a + 2d) + (a + 4d) + \cdots + (a + 22d) Simplify the equation: 12a+2d(1+2+⋯+11)=−725a 12a + 2d(1 + 2 + \cdots + 11) = -\dfrac{72}{5} a Use the formula for sum of an arithmetic series (first 11 terms): 1+2+⋯+11=11×122=66 1 + 2 + \cdots + 11 = \dfrac{11 \times 12}{2} = 66 So, the equation becomes: 12a+2d(66)=−725a 12a + 2d(66) = -\dfrac{72}{5} a 12a+132d=−725a 12a + 132d = -\dfrac{72}{5} a Solve for the relationship between aa and dd: 132d=−725a−12a 132d = -\dfrac{72}{5} a - 12a 132d=−1325a 132d = -\dfrac{132}{5} a So, a=−5d(i) a = -5d \quad \text{(i)} Second condition: ∑k=1nak=0⇒Sn=0 \sum_{k=1}^{n} a_k = 0 \quad \Rightarrow \quad S_n = 0 n2[2a+(n−1)d]=0 \dfrac{n}{2} [2a + (n-1)d] = 0 2a=−(n−1)d(ii) 2a = -(n-1)d \quad \text{(ii)} Combine equation (i) and (ii): Substitute a=−5d a = -5d into equation (ii): 2(−5d)=−(n−1)d 2(-5d) = -(n-1)d −10d=−(n−1)d -10d = -(n-1)d Therefore: n−1=10 n-1 = 10 n=11 n = 11 Thus, n=11 n = 11 .

Q188
Suppose that the number of terms in an A.P. is 2k,k∈N2 k, k \in N. If the sum of all odd terms of the A.P. is 40 , the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27 , then k is equal to:
A 8
B 6
C 4
D 5
Correct Answer
Option D
Solution
a1,a2,a3,……,a2k→A.P.∑r=1ka2r−1=40,∑r=1ka2r=55,a2k−a1=27k2[2a1+(k−1)2 d]=40,k2[2a2+(k−1)2 d]=55, d=272k−1a1=40k−(k−1)d=55k−kd d=15k⇒272k−1=15k⇒9k=10k−5∴k=5\begin{aligned} & a_1, a_2, a_3, \ldots \ldots, a_{2 k} \quad \rightarrow \mathrm{A.P.}\\ & \sum_{\mathrm{r}=1}^{\mathrm{k}} \mathrm{a}_{2 \mathrm{r}-1}=40, \sum_{\mathrm{r}=1}^{\mathrm{k}} \mathrm{a}_{2 \mathrm{r}}=55, \mathrm{a}_{2 \mathrm{k}}-\mathrm{a}_1=27 \\ & \frac{\mathrm{k}}{2}\left[2 \mathrm{a}_1+(\mathrm{k}-1) 2 \mathrm{~d}\right]=40, \frac{\mathrm{k}}{2}\left[2 \mathrm{a}_2+(\mathrm{k}-1) 2 \mathrm{~d}\right]=55, \\ & \mathrm{~d}=\frac{27}{2 \mathrm{k}-1} \\ & \mathrm{a}_1=\frac{40}{\mathrm{k}}-(\mathrm{k}-1) \mathrm{d}=\frac{55}{\mathrm{k}}-\mathrm{kd} \\ & \mathrm{~d}=\frac{15}{\mathrm{k}} \Rightarrow \frac{27}{2 \mathrm{k}-1}=\frac{15}{\mathrm{k}} \Rightarrow 9 \mathrm{k}=10 \mathrm{k}-5 \\ & \therefore \mathrm{k}=5 \end{aligned}
Q189
Let ana_n be the nthn^{th} term of an A.P. If Sn=a1+a2+a3+…+an=700S_n = a_1 + a_2 + a_3 + \ldots + a_n = 700, a6=7a_6 = 7 and S7=7S_7 = 7, then ana_n is equal to :
A 65
B 56
C 70
D 64
Correct Answer
Option D
Solution

Sum of n terms of an AP (SnS_n):

Sn=n2[2a+(n−1)d]S_n = \frac{n}{2} [2a + (n-1)d]

where 'a' is the first term and 'd' is the common difference. nth term of an AP (ana_n):

an=a+(n−1)da_n = a + (n-1)d

Given Information: Sn=700S_n = 700 (The sum of the first n terms is 700) a6=7a_6 = 7 (The 6th term is 7) S7=7S_7 = 7 (The sum of the first 7 terms is 7) Steps to Solve: Use S7S_7 to find a relationship between 'a' and 'd':

S7=72[2a+(7−1)d]=7S_7 = \frac{7}{2} [2a + (7-1)d] = 7
72[2a+6d]=7\frac{7}{2} [2a + 6d] = 7
2a+6d=22a + 6d = 2
a+3d=1a + 3d = 1

(Equation 1) Use a6a_6 to find another relationship between 'a' and 'd':

a6=a+(6−1)d=7a_6 = a + (6-1)d = 7
a+5d=7a + 5d = 7

(Equation 2) Solve the system of equations (Equation 1 and Equation 2) to find 'a' and 'd': Subtract Equation 1 from Equation 2:

(a+5d)−(a+3d)=7−1(a + 5d) - (a + 3d) = 7 - 1
2d=62d = 6
d=3d = 3

Substitute d = 3 into Equation 1:

a+3(3)=1a + 3(3) = 1
a+9=1a + 9 = 1
a=−8a = -8

Find 'n' using Sn=700S_n = 700:

Sn=n2[2a+(n−1)d]=700S_n = \frac{n}{2} [2a + (n-1)d] = 700
n2[2(−8)+(n−1)(3)]=700\frac{n}{2} [2(-8) + (n-1)(3)] = 700
n[−16+3n−3]=1400n[-16 + 3n - 3] = 1400
n[3n−19]=1400n[3n - 19] = 1400
3n2−19n−1400=03n^2 - 19n - 1400 = 0

Solve the quadratic equation for 'n': We can use the quadratic formula:

n=−b±b2−4ac2an = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
n=19±(−19)2−4(3)(−1400)2(3)n = \frac{19 \pm \sqrt{(-19)^2 - 4(3)(-1400)}}{2(3)}
n=19±361+168006n = \frac{19 \pm \sqrt{361 + 16800}}{6}
n=19±171616n = \frac{19 \pm \sqrt{17161}}{6}
n=19±1316n = \frac{19 \pm 131}{6}

We have two possible values for n:

n=19+1316=1506=25n = \frac{19 + 131}{6} = \frac{150}{6} = 25
n=19−1316=−1126=−563n = \frac{19 - 131}{6} = \frac{-112}{6} = -\frac{56}{3}

(Since 'n' must be a positive integer, we discard this solution) So,

n=25n = 25

Find ana_n (which is a25a_{25}):

a25=a+(25−1)da_{25} = a + (25-1)d
a25=−8+(24)(3)a_{25} = -8 + (24)(3)
a25=−8+72a_{25} = -8 + 72
a25=64a_{25} = 64

Therefore, an=64a_n = 64. The correct answer is Option D.

Q190
If the sum of the first 20 terms of the series 4⋅14+3⋅12+14+4⋅24+3⋅22+24+4⋅34+3⋅32+34+4⋅44+3⋅42+44+…⋅\dfrac{4 \cdot 1}{4+3 \cdot 1^2+1^4}+\dfrac{4 \cdot 2}{4+3 \cdot 2^2+2^4}+\dfrac{4 \cdot 3}{4+3 \cdot 3^2+3^4}+\dfrac{4 \cdot 4}{4+3 \cdot 4^2+4^4}+\ldots \cdot is mn\dfrac{\mathrm{m}}{\mathrm{n}}, where m and n are coprime, then m+n\mathrm{m}+\mathrm{n} is equal to :
A 423
B 421
C 422
D 420
Correct Answer
Option B
Solution
Sn=∑r=1n4r4+3r2+r4=2∑r=1n2r(r2+2)2−r2=2∑r=1n(r2+2+r)−(r2+2−r)(r2+2+r)(r2+2−r)=2∑r=1n(1r2+2−r−1r2+2+r)S20=2[(12−14)+(14−18)+…]=2(12−1202+2+20)=2(12−1422)=2(422−2422×2)=420422=210211=mnm+n=421\begin{aligned} & S_n=\sum_{r=1}^n \frac{4 r}{4+3 r^2+r^4} \\ & =2 \sum_{r=1}^n \frac{2 r}{\left(r^2+2\right)^2-r^2}=2 \sum_{r=1}^n \frac{\left(r^2+2+r\right)-\left(r^2+2-r\right)}{\left(r^2+2+r\right)\left(r^2+2-r\right)} \\ & =2 \sum_{r=1}^n\left(\frac{1}{r^2+2-r}-\frac{1}{r^2+2+r}\right) \\ & S_{20}=2\left[\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+\ldots\right] \\ & \quad=2\left(\frac{1}{2}-\frac{1}{20^2+2+20}\right) \\ & \quad=2\left(\frac{1}{2}-\frac{1}{422}\right) \\ & \quad=2\left(\frac{422-2}{422 \times 2}\right)=\frac{420}{422}=\frac{210}{211}=\frac{m}{n} \\ & m+n=421 \end{aligned}
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