Sequences and Series Test

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Sequences and Series Test - 51

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Question 1
1 / -0

Sum the following series to n terms: $$1+3+6+10+15+...$$

A
$$\dfrac{n}{6}(n-1)(n-2)$$

B
$$\dfrac{n}{3}(n+1)(n-1)$$

C
$$\dfrac{n}{6}(n+1)(n+2)$$

D
None of these

Solution

Gives series $$1+3+6+10+15+.........$$

The sequence of the difference between the successive terms of the series is $$2, 3, 4, 5 +......$$

Clearly it is an $$A.P$$ with common difference $$1$$

Let $$T_{n}$$ be the $$n^{th}$$ term & $$S_{n}$$ denote the sum of $$n$$ terms of the given series

Then $$S_{n}=1+3+6+10+15+.....+T_{n-1}+T_{n} ........ (1)$$

Also $$S_{n}1+3+6+10+....... T_{n-1}+T_{n} ...... (2)$$

Subtracting $$(2)$$ from $$(1)$$, we get

$$0=1+[2+3+4+5+..... T_{n}+T_{n-1}]-T_{n}$$

$$T_{n}=1+\dfrac{(n-1)}{2}[2\times 2+(n-1-1)\times 1]$$

$$T_{n}=1+\dfrac{(n-1)}{2}(n+2)=\dfrac{2+n^{2}+2n-n-2}{2}$$

$$T_{n}=\dfrac{n^{2}+n}{2}$$

$$S_{n}=\displaystyle\sum_{k=1}^{n}T_{k}=\sum_{k=1}^{n}\left(\dfrac{k^{2}+k}{2}\right)=\dfrac{1}{2}\sum_{k=1}^{n}k^{2}+\dfrac{1}{2}\sum_{k=1}^{n}k$$

$$S_{n}=\dfrac{1}{2}\left[\dfrac{n(n+1)(2n+1)}{6}+\dfrac{n(n+1)}{2}\right]$$

$$S_{n}=\dfrac{1}{2}\left[n(n+1)\left[\dfrac{2n+1+3}{6}\right]\right]$$

$$S_{n}=\dfrac{n}{6}(n+1)(n+2)$$
Question 2
1 / -0

Find the sum of the following series to n terms:
$$1\times 2+2\times 3+3\times 4+4\times 5+...$$

A
$$\dfrac{n}{4}(n-1)(n+2)$$

B
$$\dfrac{n}{3}(n-1)(n-2)$$

C
$$\dfrac{n}{2}(n-1)(n+1)$$

D
$$\dfrac{n}{3}(n+1)(n+2)$$

Solution

Let $$T_n$$ be the $$n^{th}$$ term of the given series, Then

$$T_n=(n^{th}\ term\ of\ 1,2,3....)\times (n^{th}\ term\ of\ 2,3,4....)$$

$$=[1+(n-1)\times 1]. [2+(n-1)\times 1]$$

$$=[1+n-1][2+n-1]$$

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

Let $$S_n$$ denotes sum of $$n$$ term of the given series

$$S_n= \displaystyle \sum_{n=1}^{n}T_n =\displaystyle \sum_{n=1}^{n} (n^2 +n)=\displaystyle \sum_{n=1}^{n}n^2 +\displaystyle \sum_{n=1}^{n}n$$

$$=\dfrac {n(n+1)(2n+1)}{6}+\dfrac {n(n+1)}{2}=\dfrac {n(n+1) (2n+1)+3n(n+1)}{6}$$

$$=\dfrac {n(n+1)(2n+4)}{6}=\dfrac {n(n+1)\times 2(n+2)}{6}$$

$$S_n =\dfrac {n}{3}(n+1)(n+2)$$
Question 3
1 / -0

Find the sum of the following series to n terms:
$$1+(1+2)+(1+2+3)+(1+2+3+4)+...$$

A
$$\dfrac{n}{6}(n+1)(n+2)$$

B
$$\dfrac{n}{6}(n-1)(n-2)$$

C
$$\dfrac{n}{6}(n+1)(n-1)$$

D
None of these

Solution

Let $$T_n$$ be the $$n^{th}$$ term of the given series,

Then,

$$T_n =1+2+3+.....+n$$

$$=\dfrac {n}{2}[2\times 1+(n-1)\times 1]$$

$$=\dfrac {n}{2} [2+n-1]=\dfrac {n}{2} (n+1)$$

$$=\dfrac {n^2}{2}+\dfrac {n}{2}$$

Let $$S_n$$ denoted sum of $$n$$ terms of the given series

$$S_n =\displaystyle \sum_{K=1}^nT_K=\displaystyle \sum_{K=1}^n \left [\dfrac {K^2}{2}+K/2\right] $$

$$S_n =\dfrac {1}{2} \displaystyle \sum_{K=1}^nK^2 +\dfrac {1}{2}\displaystyle \sum_{K=1}^nK$$

$$S_n=\dfrac {1}{2} \left [\dfrac {n(n+1) (2n+1)}{6} \right] +\dfrac {1}{2} \left [\dfrac {n(n+1)}{2}\right]$$

$$S_n=\dfrac {n(n+1)(2n+1)+3n(n+1)}{12}$$

$$S_n =\dfrac {n(n+1)[2n+4]}{12}$$

$$S_n=\dfrac {n(n+1)}{12}\times 2(n+2)$$

$$S_n=\dfrac {n}{6}(n+1)(n+2)$$
Question 4
1 / -0

The next number in the pattern 62, 37, 12 ____ is

A
25

B
13

C
0

D
-13

Solution

Option (b) is correct; here, the given series has a common difference of +25.
$$\begin{array}{l}\Rightarrow-62+25=-37,-37+25=-12 \\\therefore-12+25=13\end{array}$$
Question 5
1 / -0

If AM of two numbers a and 17 is 15. Then a is ___

A
15

B
13

C
17

D
None of the above

Solution
Question 6
1 / -0

Arithmetic mean of two numbers a+d and a-d is

A
$$a$$

B
$$b$$

C
Cannot be determined

D
None of the above

Solution
Question 7
1 / -0

Find the sum of Arithmetic means of $$3 , 9$$ and $$12 , 8$$.

A
$$6$$

B
$$9$$

C
$$20$$

D
$$16$$

Solution
Question 8
1 / -0

Insert an arithmetic mean between $$7$$ and $$21$$

A
$$10$$

B
$$14$$

C
$$28$$

D
$$30$$
Question 9
1 / -0

The arithmetic mean of $$4$$ and $$14$$ is

A
$$9$$

B
$$18$$

C
$$10$$

D
$$4$$

Solution
Question 10
1 / -0

Sum of the series $$S=1^2-2^2+3^2-4^2+.....-2002^2+2003^2$$ is

A
$$2007006$$

B
$$1005004$$

C
$$2000506$$

D
$$1005040$$

Solution

$$S = {1}^{2} - {2}^{2} + {3}^{2} - {4}^{2} + ...... - {2002}^{2} + {2003}^{2}$$

$$\displaystyle \Rightarrow S = \sum_{k = 1}^{2003}{{\left( -1 \right)}^{k + 1}{k}^{2}}$$

$$\displaystyle \Rightarrow S = \sum_{k = 1}^{1002}{{\left( 2k - 1 \right)}^{2}} - \sum_{k = 1}^{1001}{{\left( 2k \right)}^{2}}$$

$$\displaystyle \Rightarrow S = {\left[ \left( 2 \times 1002 \right) - 1 \right]}^{2} + \sum_{k = 1}^{1001}{\left[ {\left( 2k \right)}^{2} - {\left( 2k \right)}^{2} - 4k + 1 \right]}$$

$$\displaystyle \Rightarrow S = {\left( 2003 \right)}^{2} - \sum_{k = 1}^{1001}{\left( 4k - 1 \right)} = {\left( 2003 \right)}^{2} - 4 \times \cfrac{1001 \left( 1001 + 1 \right)}{2} + 1001 = 2007006$$