11th Standard Syllabus & Materials

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11th Maths Unit 5 Binomial Theorem, Sequences and Series One Mark Question with Answer Key

Published on: 27/08/2019

Binomial Theorem, Sequences and Series

Download Tamil Nadu 11th Standard Maths question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.

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Questions:

Multiple Choice Question

10 x 1 = 10

The value of \({ 9 }^{ \frac { 1 }{ 3 } }\) ,\({ 9 }^{ \frac { 1 }{ 9 } }\)\({ 9 }^{ \frac { 1 }{ 27}}\),\(\infty \) is ______________

(a)

(b)

(c)

(d)

none of these

The nth term of a G.P is 128 and the sum of its n terms is 225. If its common ratio is 2, then its first term is ______________

(a)

(b)

(c)

(d)

none of these

If in an infinite G. P. first term is equal to 10 times the sum of all successive terms, then its common ratio is ______________

(a)

\(\frac { 1 }{ 10 } \)

(b)

\(\frac { 1 }{ 11 } \)

(c)

\(\frac { 1 }{ 9 } \)

(d)

\(\frac { 1 }{ 20 } \)

The first and last term of an A.P. are 1 and 11. If the sum of its terms is 36, then the number of terms will be ______________

(a)

(b)

(c)

(d)

The term without x in \({ \left( 2x-\frac { 1 }{ 2{ x }^{ 2 } } \right) }^{ 12 }\) is ______________

(a)

495

(b)

-495

(c)

-7920

(d)

7920

If the sum of n terms of an A. P. be 3n² - n and its common difference is 6, then its first term is ______________

(a)

(b)

(c)

(d)

The sum of an infinite GP is 18. If the first term is 6, the common ratio is

(a)

\(\frac { 1 }{ 3 } \)

(b)

\(\frac { 2 }{ 3 } \)

(c)

\(\frac { 1 }{ 6 } \)

(d)

\(\frac { 3 }{ 4 } \)

The n^thterm of the sequence 1, 2, 4, 7, 11,... is

(a)

n³+ 3n²+ 2n

(b)

n³- 3n²+ 3n

(c)

\(\frac { n(n+1)(n+2) }{ 3 } \)

(d)

\(\frac { { n }^{ 2 }-n+2 }{ 2 } \)

The HM of two positive numbers whose AM and GM are 16, 8 respectively is

(a)

(b)

(c)

(d)

10.

If a, 8, b are in AP, a, 4, b are in GP, and if a, x, b are in HP then x is

(a)

(b)

(c)

(d)

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Answers Key:

Multiple Choice Question

10 x 1 = 10

(b)
3

(a)
1

(b)
\(\frac { 1 }{ 11 } \)

(b)
6

(d)
7920

(a)
2

\(a=6, S_{\infty}=18, r=? \)
\(S_{\infty} =\frac{a}{1-r}=\frac{6}{1-r}=18 \)
\(6 =18-18 r \)
\(18 r =12 \)
\(r =\frac{2}{3} \)

\(n^{\text {th }} \text { term is } \frac{n^{2}-n+2}{2}\)

\(\mathrm{AM}=16, \quad \mathrm{GM}=8, \quad \mathrm{HM}=?\)
\(\frac{a+b}{2}=16 \Rightarrow a+b=32\)
\(\sqrt{a b}=8 \Rightarrow a b=64\)
\(\therefore \mathrm{HM} =\frac{2 a b}{a+b} \)
\(=\frac{2(64)}{32}= 4 \)

10.

\(a+b= 16, a b=16, x=\frac{2 a b}{a+b} \)
\(x=\frac{2 \times 16}{16}\)
\(x=2\)