(Each question 2 marks)

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 (Each question 2 marks)