2 Marks Questions

Question 512 Marks

Find: 18th term of the A.P. $\sqrt{2},\ 3\sqrt{2},\ 5\sqrt{2},\ ...$

Answer

To find 18th term of A.P. $\sqrt{2},3\sqrt{2},5\sqrt{2},\ ...$
Here, 1st term $\text{a}_1=\sqrt{2}$
and d = cpmmon difference $=2\sqrt{2}$
$\therefore\text{a}_\text{n}=\text{a}_1+(\text{n}-1)\text{d}$
$\text{a}_{18}=\sqrt{2}+2\sqrt{2}(17)=35\sqrt{2}$

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Question 522 Marks

Find the sum of the following series:
$(\text{a}-\text{b})^2+(\text{a}^2+\text{b}^2)+(\text{a}+\text{b})^2+\ ...\ +[(\text{a}+\text{b}^2)+6\text{ab}]$

Answer

$(\text{a}-\text{b})^2+(\text{a}^2+\text{b}^2)+(\text{a}+\text{b})^2+\ ...\ +[(\text{a}+\text{b}^2)+6\text{ab}]$
Let number of terms be n
Then,
$\text{a}_\text{n}=(\text{a}+\text{b}^2)+6\text{ab}$
$\Rightarrow(\text{a}+\text{b})^2+(\text{n}-1)(2\text{nd})=(\text{a}+\text{b})^2+6\text{ab}$
$\Rightarrow\text{a}^2+\text{b}^2-2\text{ab}+2\text{abn}-2\text{ab}=\text{a}^2+\text{b}^2+2\text{ab}+6\text{ab}$
then,
$\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
$\text{s}_6=\frac{6}{2}[\text{a}^2+\text{b}^2-2\text{ab}+\text{a}^2+\text{b}^2+\text{ab}+6\text{ab}]$
$=6[\text{a}^2+\text{b}^2+3\text{ab}]$

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Question 532 Marks

The first term of an A.P. is 5, the common difference is 3 and last term is 80; find the number of terms.

Answer

Given: $\text{a}=5$
$\text{d}=3$
$\text{a}_\text{n}=$ last there be n terms
$\therefore\text{a}_\text{n}=80=\text{a}+(\text{n}-1)\text{d}$
$80=5+(\text{n}-1)3$
$\Rightarrow\text{n}=26$
$\therefore$ Thus, thre are 26 term in the given sequence.

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Question 542 Marks

Find the first four terms of the sequence defined by $\text{a}_1=3$ and $\text{a}_\text{n}=3\text{a}_{\text{n}-1}+2$ for all $\text{n}>1.$

Answer

$\text{a}_\text{n}=3\text{a}_{\text{n}-1}+2$
$\text{a}_1=3$
$\text{a}_1=\text{3a}_{2-1}+2=\text{3a}_{2-1}+2=3(3)+2=11$
$\text{a}_3=\text{3a}_{3-1}+2=3\text{a}_2+2=3(11)+2=35$
$\text{a}_4=\text{3a}_{4-1}+2=\text{3a}_3+2=3(35)+2=107$
First four terms of the sequence are 3, 11, 35 and 107.

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Question 552 Marks

If $n$ A.M.$s$ are inserted between two numbers, prove that the sum of the means equidistant from the beginning and the end is constant.

Answer

Let $A_1, A_2......A_n $ be n A.M.s between two numbers a and b.
Then, $a, A_1, A_2......A_n,$ b are in A.P. with common difference, $\text{d}=\frac{\text{b}-\text{a}}{\text{n}+1}.$
$\therefore\text{A}_1+\text{A}^2+\ ......+\text{A}_\text{n}=\frac{\text{n}}{2}[\text{A}_1+\text{A}_\text{n}]$
$=\frac{\text{n}}{2}[\text{A}_1-\text{d}+\text{A}_\text{n}+\text{d}]$
$=\frac{\text{n}}{2}[\text{a}+\text{b}]$
$=\text{n}\times\big[\frac{\text{a}+\text{b}}{2}\big]$
$=$ A.M. between $a$ and $b,$ which is constant.

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Question 562 Marks

Find the $12^{th}$ term from the end of the following arithmetic progressions,

$3, 5, 7, 9, ... 201$

Answer

A.p. is $3, 5, 7, 9, ..., 201.$
Here,
$\text{a}=3$
$\text{d}=2$
$n^{th}$ term from the end is $\text{l}-(\text{n}-1)\text{d}$
i.e. $201-(\text{n}-1)2$ or $203-2\text{n}\ .....{(1)}$
12th term from end is
$203-2(12)=179$

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Question 572 Marks

Find the sum of all integers between $50$ and $500$, which are divisible by $7.$

Answer

The series of integers by $7$ between $50$ and $500$ are
$56, 63, 70, ..., 497$
Let the number of terms be n then, $n^{th}$ term $= 497$
$\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$\Rightarrow497=56+(\text{n}-1)7$
$\Rightarrow\text{n}=64$
The sum $\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
$\Rightarrow\text{s}_{64}=\frac{64}{2}[56+497]$
$=32\times553$
$=17696$

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Maths 2 Marks Questions