Arithmetic Progressions Test

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Arithmetic Progressions Test - 33

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Question 1
1 / -0

If the sum of all the terms of an A.P.: $$25, 22, 19, ......,$$ $$t_n$$ is $$116,$$ then $$n$$ is

A
$$8$$

B
$$4$$

C
$$2$$

D
$$16$$

Solution

Let number of terms be $$n$$

$$ \therefore \displaystyle \frac{n}{2} [50 + (n - 1) \times (-3)] = 116$$

$$\implies \displaystyle \frac 12[50n-3n^2+3n]=116$$

$$ \implies 3n^2 - 53n + 232 = 0 \implies n = \dfrac{29}{3} \ or \ 8 $$

Hence, $$n = 8$$
Question 2
1 / -0

In an A.P., if the common difference is $$2$$ and sum up to $$n$$ terms is $$49$$ and $$7^{th}$$ term is $$13$$, then value of $$n$$ is equal to?

A
$$0$$

B
$$5$$

C
$$7$$

D
$$13$$

Solution

Given: $$a_{7}=a_{1}+6d$$
It is given that the common difference is 2.
Therefore
$$a_{7}=a_{1}+12$$
$$13=a_{1}+12$$
$$a_{1}=1$$.

Now sum of n terms is given by:
$$S_{n}=\dfrac{n}{2}[2a_{1}+(n-1)d]$$

$$49=\dfrac{n}{2}[2+(n-1)2]$$

$$49=n(1+(n-1))$$

$$49=n^{2}$$

$$n=7$$
Hence the value of n is 7.
Question 3
1 / -0

In an A.P., if $$a = 3.5$$, $$d = 0$$, $$n = 101$$, then $$a_n$$ will be

A
$$0$$

B
$$3.5$$

C
$$103.5$$

D
$$104.5$$

Solution

$$a = 3.5, d=0$$ and $$n= 101$$
$$a_n = a + (n-1)\times d$$
$$a_n = 3.5 + 100\times0$$
$$a_n= 3.5$$
Question 4
1 / -0

For an A.P. $$a_1,a_2,...a_n$$ and $$a_1=\frac{5}{2},a_{10}=16$$.
If sum of $$n$$ terms is $$110$$ then $$n$$ equals

A
$$9$$

B
$$10$$

C
$$11$$

D
$$12$$

Solution

$$a_{1}=\dfrac{5}{2}$$
Let the common difference be d.
Then
$$a_{10}=a_{1}+9d$$

$$16=\dfrac{5}{2}+9d$$

$$\dfrac{27}{2}=9d$$

$$d=\dfrac{3}{2}$$

Hence

$$S_{n}=\dfrac{n}{2}(2(\dfrac{5}{2})+(n-1)\dfrac{3}{2})$$

$$=\dfrac{n}{4}(10+3(n-1))$$

$$=\dfrac{n}{4}(7+3n)$$

Now $$S_{n}=110$$
Hence
$$\dfrac{n}{4}(7+3n)=110$$

$$n(7+3n)=440$$

$$3n^{2}+7n-440=0$$

$$n=11$$ and $$n=\dfrac{-40}{3}$$
Since
$$n=\dfrac{-40}{3}$$ is not possible, hence $$n=11$$
Question 5
1 / -0

If $$8^{th}$$ term of an A.P is $$15$$, then the sum of $$15$$ terms is

A
$$15$$

B
$$0$$

C
$$225$$

D
$$\dfrac{225}2$$

Solution

$$a_{8}=a_{1}+7d$$
Hence
$$a_{1}+7d=15$$ ...(i)
$$S_{15}=\dfrac{n}{2}[2a_{1}+(15-1)d]$$

$$S_{15}=\dfrac{15}{2}[2a_{1}+14d]$$

$$=15(a_{1}+7d)$$

$$=15(15)$$ ...from i
$$=225$$.
Question 6
1 / -0

The $$11^{th}$$ term of the series $$ \displaystyle -5,\frac{-5}{2},0,\frac{5}{2},\cdots$$ is

A
$$20$$

B
$$30$$

C
$$40$$

D
$$50$$

Solution

$$a = -5, d=\dfrac{5}{2}$$ and $$n= 11$$.
$$a_{11} = -5 + (11-1)\times d$$
$$a_{11} = -5 + 10\times\dfrac{5}{2}$$
$$a_{11}= 20$$
Question 7
1 / -0

If the first term of an AP is $$5$$ and the common difference is $$-2,$$ then the sum of the first $$6$$ terms is

A
$$0$$

B
$$5$$

C
$$6$$

D
$$15$$

Solution

Sum of an AP is: $${\dfrac{n}{2}}\times[{2a+(n-1)d}]$$
$$a = 5, d= -2, n=6$$

Sum $$={\dfrac{6}{2}}\times[{(2\times5)+(6-1)\times(-2)}]$$

$$= 3\times(10-10)$$
$$= 0$$
Question 8
1 / -0

In an AP if $$a = 1$$, $${ a }_{ n }=20$$ and $${ S }_{ n }=399$$, then $$n$$ is

A
$$19$$

B
$$21$$

C
$$38$$

D
$$42$$

Solution

$$a = 1$$
$$a_n = 20$$
$$S_n= 399$$
Then,
$$\Rightarrow a_n=a+(n-1)d$$
$$\Rightarrow 20=1 + (n-1)d$$
$$\Rightarrow 19=(n-1)d$$
Also given,
$$S_n = \dfrac{n}{2} \times$$$$ [2a+(n-1)d]$$
Substituting values,
$$399\times 2=n( 2+19)$$
$$\Rightarrow 798$$ = $${n\times21}$$
$$n = 38$$
Question 9
1 / -0

If the common difference of an AP is $$5,$$ then what is $$a_{18}-a_{13}$$ ?

A
$$5$$

B
$$20$$

C
$$25$$

D
$$30$$

Solution

Hint: $$n^{th }$$ term of an A.P. is $$a_n=a+(n-1)d$$ where $$a$$ is first term and $$d$$ is common difference.

Given:
value of common difference $$,d=5$$

Step 1 : Find the value of $$18^{th}$$ term i.e, $$a_{18}$$ by replacing $$n$$ with $$18$$ in the $$n^{th}$$ term of the A.P
$$\Rightarrow a_n=a+(n-1)d$$
$$\Rightarrow a_{18}=a+(18-1)d$$
$$\therefore a_{18}=a+17d$$

Step 2 : Find the value of $$13^{th}$$ term i.e, $$a_{13}$$ by replacing $$n$$ with $$13$$ in the $$n^{th}$$ term of the A.P
$$\Rightarrow a_n=a+(n-1)d$$
$$\Rightarrow a_{13}=a+(13-1)d$$
$$\therefore a_{13}=a+12d$$

Step 3 : Find the difference between $$a_{18}$$ and $$a_{13}$$
$$a_{18}-a_{13}$$
$$=[a+17d]-[a+12d]$$
$$=a+17d-a-12d$$
$$=5d\\=5\times5\\=25$$

Final step : Hence, $$a_{18}-a_{13}=25$$
Question 10
1 / -0

Which term of the AP, $$21, 42, 63, 84,...$$ is $$210?$$

A
$$9^{th}$$

B
$$10^{th}$$

C
$$11^{th}$$

D
$$12^{th}$$

Solution

First term of the AP, $$a=21$$
Common difference $$d=21$$
$${n}^{th}$$ term is $$210$$
Then,
$$210 = a + (n-1)d$$
$$210= 21 + (n-1)\times21$$
$$210= 21n$$
Therefore, $$n=10$$
Hence, 210 is $${ 10 }^{ th }$$ term of the series.

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Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 33

Result

Arithmetic Progressions Test - 33

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SHARING IS CARING

Question 1

$$8$$

$$4$$

$$2$$

$$16$$

Solution

Question 2

$$0$$

$$5$$

$$7$$

$$13$$

Solution

Question 3

$$0$$

$$3.5$$

$$103.5$$

$$104.5$$

Solution

Question 4

$$9$$

$$10$$

$$11$$

$$12$$

Solution

Question 5

$$15$$

$$0$$

$$225$$

$$\dfrac{225}2$$

Solution

Question 6

$$20$$

$$30$$

$$40$$

$$50$$

Solution

Question 7

$$0$$

$$5$$

$$6$$

$$15$$

Solution

Question 8

$$19$$

$$21$$

$$38$$

$$42$$

Solution

Question 9

$$5$$

$$20$$

$$25$$

$$30$$

Solution

Question 10

$$9^{th}$$

$$10^{th}$$

$$11^{th}$$

$$12^{th}$$

Solution

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