Arithmetic Progressions Test

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Arithmetic Progressions Test - 62

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Weekly Quiz Competition

Question 1
1 / -0

In an arithmetic progression, if $${ S }_{ n }=n(5+3n)\quad and\quad { t }_{ n }=32$$, then the value of n is
[Note: $${ S }_{ n }$$ and $${ t }_{ n }$$ denote the sum of first n terms and $${ n }^{ th }$$ term of arithmetic progression respectively.]

A
4

B
5

C
6

D
7

Solution
Question 2
1 / -0

If the sum of the three consecutive terms of A.P is $$48$$ and product of the first the last is $$252$$, and d=........

A
2

B
3

C
4

D
16

Solution
Question 3
1 / -0

The sum of five consecutive odd numbers is 295. The smallest number is

A
51

B
53

C
55

D
57
Question 4
1 / -0

Let a, b, c be three distinct positive numbers which are in G.P, if $${ log }_{ c }a,{ log }_{ b }c,{ log }_{ a }b$$ are in A.P., then the common difference of the A.P is

A
$$\cfrac { 1 }{ 2 } $$

B
$$\cfrac { 3 }{ 2 } $$

C
$$\cfrac { 5 }{ 2 } $$

D
$$\cfrac { 7 }{ 2 } $$
Question 5
1 / -0

The number of common terms in two AP's 2, 5, 8, 11,..., 179 and 3, 5, 7, 9, .. 101 are

A
15

B
16

C
17

D
18

Solution
Question 6
1 / -0

The sum of n terms of an AP is $$(3n^2 + 5n).$$ Which of its terms is 164?

A
28th

B
27th

C
26th

D
29th

Solution

Given sum of n terms is,

$${ S }_{ n }=3{ n }^{ 2 }+5n$$

Put n=1, we get,
$${ S }_{ 1 }=a_{ 1 }=3{ \left( 1 \right) }^{ 2 }+5\left( 1 \right) $$
$$\therefore a_{ 1 }=3+5$$
$$\therefore a_{ 1 }=8$$

Put n=2 in above equation, we get,
$${ S }_{ 2 }={ a }_{ 1 }+{ a }_{ 2 }=3{ \left( 2 \right) }^{ 2 }+5\left( 2 \right) $$
$$\therefore 8+{ a }_{ 2 }=\left( 3\times 4 \right) +10$$
$$\therefore 8+{ a }_{ 2 }=22$$
$$\therefore { a }_{ 2 }=14$$

$$d=14-8$$
$$\therefore d=6$$

Now, Given $${ a }_{ n }=164$$
$${ a }_{ n }={ a }_{ 1 }+\left( n-1 \right) d$$

$$\therefore 164=8+\left( n-1 \right) 6$$

$$\therefore 156=6\left( n-1 \right) $$

$$\therefore n-1=26$$

$$\therefore n=27$$
Question 7
1 / -0

In an AP, the correct relation is

A
$$a_{n - 5} = a_{n - 4} + d$$

B
$$a_{n - 5} = a_{n - 6} + d$$

C
$$a_{n - 5} = a_{n} + d$$

D
$$a_{n + 5} = a_{n - 5} - d$$

Solution

We know that, in an A.P., the common difference $$(d)$$ is the difference between any two consecutive terms.

If we take one of the terms as $$a_{n-6}$$, we will need either $$a_{n-7}$$ or $$a_{n-5}$$ for finding the common difference.

If we take $$a_{n-6}$$ and $$a_{n-5}$$, then,
$$d={ a }_{ n-5 }-{ a }_{ n-6 }$$

$$\therefore \ { a }_{ n-5 }={ a }_{ n-6 }+d$$

Hence, option B is correct.
Question 8
1 / -0

In an A.P $$a_{n + 5} = 50$$ and $$a_{n + 1} = 38$$ then common difference

A
$$3$$

B
$$2$$

C
$$3n$$

D
$$2n$$

Solution

nth term of A.P. is given by,
$${ a }_{ n }=a+\left( n-1 \right) d$$ (1)

1) Put $$n=n+5$$ in (1), we get,
$${ a }_{ n+5 }=a+\left( n+5-1 \right) d$$

$$\therefore 50=a+\left( n+4 \right) d$$

$$\therefore \dfrac { 50-a }{ d } =n+4$$

$$\therefore n=\dfrac { 50-a }{ d } -4$$ (2)

2) Put $$n=n+1$$ in (1), we get,
$${ a }_{ n+1 }=a+\left( n+1-1 \right) d$$

$$\therefore 38=a+nd$$

$$\therefore n=\dfrac { 38-a }{ d } $$ (3)

From (2) and (3), we get,
$$\dfrac { 50-a }{ d } -4=\dfrac { 38-a }{ d }$$

$$\dfrac { 50-a }{ d } -\dfrac { 38-a }{ d } =4$$

$$\therefore \dfrac { 50-a-\left( 38-a \right) }{ d } =4$$

$$\therefore \dfrac { 50-a-38+a }{ d } =4$$

$$\therefore \dfrac { 50-38 }{ d } =4$$

$$\therefore \dfrac { 12 }{ d } =4$$

$$\therefore d=3$$
Question 9
1 / -0

How many terms of the AP 6, 12, 18, 24,.... must be take to make the sum 816?

A
16

B
18

C
14

D
22

Solution

Given,
$${ a }_{ 1 }=6$$
$$d=12-6=6$$

Since $$ S_n =816$$

Sum of n terms of A.P. is given as,
$$ S_n =\dfrac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] $$

$$\therefore 816=\dfrac { n }{ 2 } \left[ \left( 2\times 6 \right) +\left( n-1 \right) 6 \right] $$

$$\therefore 1632=n\left[ 12+6n-6 \right] $$

$$\therefore 1632=n\left[ 6n+6 \right] $$

$$\therefore 1632=6{ n }^{ 2 }+6n$$

$$\therefore 6{ n }^{ 2 }+6n-1632=0$$
Divide both sides by 6, we get,

$${ n }^{ 2 }+n-272=0$$

$$\therefore n=\dfrac { -1\pm \sqrt { { \left( 1 \right) }^{ 2 }-4\times 1\times \left( -272 \right) } }{ 2\times 1 } $$

$$\therefore n=\dfrac { -1\pm \sqrt { 1+1088 } }{ 2 } $$

$$\therefore n=\dfrac { -1\pm \sqrt { 1089 } }{ 2 } $$

$$\therefore n=\dfrac { -1\pm 33 }{ 2 } $$

Consider only positive root.

$$\therefore n=\dfrac { -1+33 }{ 2 } $$

$$\therefore n=\dfrac { 32 }{ 2 } $$

$$\therefore n=16$$
Question 10
1 / -0

How many terms of the $$ AP -5, \frac{-9}{2}, -4, $$ .... will give the sum $$0$$?

A
$$21$$

B
$$18$$

C
$$23$$

D
$$16$$

Solution

Given,
$${ a }_{ 1 }=-5$$
$$d=\frac { -9 }{ 2 } -\left( -5 \right) $$
$$=\frac { -9 }{ 2 } +5$$
$$=\frac { 1 }{ 2 } $$

Since $$ { { S }_{ n }= } 0$$

Thus, sum of n terms of A.P. are calculated as,
$$ { { S }_{ n }= } \frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] $$

$$\therefore 0=\frac { n }{ 2 } \left[ \left( 2\times (-5) \right) +\left( n-1 \right) \frac { 1 }{ 2 } \right] $$

$$\therefore \left( 2\times (-5) \right) +\left( n-1 \right) \frac { 1 }{ 2 } =0$$

$$-10+\frac { \left( n-1 \right) }{ 2 } =0$$

$$\frac { \left( n-1 \right) }{ 2 } =10$$

$$\left( n-1 \right) =20$$

$$\therefore n=20+1$$

$$\therefore n=21$$

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Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 62

Result

Arithmetic Progressions Test - 62

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SHARING IS CARING

Question 1

4

5

6

7

Solution

Question 2

2

3

4

16

Solution

Question 3

51

53

55

57

Question 4

$$\cfrac { 1 }{ 2 } $$

$$\cfrac { 3 }{ 2 } $$

$$\cfrac { 5 }{ 2 } $$

$$\cfrac { 7 }{ 2 } $$

Question 5

15

16

17

18

Solution

Question 6

28th

27th

26th

29th

Solution

Question 7

$$a_{n - 5} = a_{n - 4} + d$$

$$a_{n - 5} = a_{n - 6} + d$$

$$a_{n - 5} = a_{n} + d$$

$$a_{n + 5} = a_{n - 5} - d$$

Solution

Question 8

$$3$$

$$2$$

$$3n$$

$$2n$$

Solution

Question 9

16

18

14

22

Solution

Question 10

$$21$$

$$18$$

$$23$$

$$16$$

Solution

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