Arithmetic Progressions Test

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Arithmetic Progressions Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

If $$n^{th}$$ term of an AP is $$\dfrac {1}{3}(2n+1)$$, then the sum of its $$19$$ term is

A
$$131$$

B
$$132$$

C
$$133$$

D
$$134$$

Solution

Let $$T_n$$ be the $$n^{th}$$ term of an AP

$$\Rightarrow T_n=a+(n-1)d\\\ \ \ \ \ \ \ \ \ =nd+a-d$$

On comparing with the given expression it is clear that $$d=\dfrac{2}{3}$$ and $$a-d=\dfrac{1}{3}$$

$$\Rightarrow a-\dfrac{2}{3}=\dfrac{1}{3}$$

$$\Rightarrow a=1$$

We know that sum of first $$n$$ terms, $$S_n = \dfrac{n}{2} (2a+(n-1)d)$$, where $$a$$ & $$d$$ are the first term amd common difference of an AP.

$$\therefore S_{19}=\dfrac {19}{2}\left [ 2+(18)\dfrac{2}{3} \right ]\\=133$$

Hence, option $$(C)$$ is correct.
Question 2
1 / -0

Find the common difference of an AP. whose first term is $$100$$ and the sum of whose first six terms is five times the sum of the next six terms.

A
$$10$$

B
$$-10$$

C
$$5$$

D
$$-5$$

Solution

Here $$a=100$$

Let difference is $$d$$.
$$\Rightarrow a_1+a_2+a_3+a_4+a_5+a_6=5(a_7+a_8+a_9+a_{10}+a_{11}+a_{12})$$
So by the formula, $$S_n = \dfrac{n}{2} (a+l) $$, where $$a$$ &$$l$$ are the first and last term of an AP, we have
$$ 6\left(\dfrac{a_1+a_6}{2} \right)=5\times 6\left(\dfrac{a_7+a_{12}}{2}\right)$$

$$\Rightarrow a_1+a_6=5(a_7+a_{12})$$

$$\Rightarrow a+a+5d=5(a+6d+a+11d)$$

$$\Rightarrow 2a+5d=10a+85d$$

$$\Rightarrow 80d=-8a$$

$$\Rightarrow d=\dfrac{-a}{10}\Rightarrow \dfrac{-100}{10}\Rightarrow -10$$
Question 3
1 / -0

Which term of AP : $$3, 15, 27, 39,..$$ will be $$132$$ more than its $$54th$$ term?

A
t$$_{65}$$

B
t$$_{64}$$

C
t$$_{60}$$

D
None of these

Solution

Clearly, the given Sequence is an AP with first term $$a=3$$ and common difference $$d=15-3=12$$.
Let $$nth$$ term be the $$132$$ more than $$54th$$ term of the given AP.
$$\Rightarrow a_n=a_{54}+132$$
$$\Rightarrow a+(n-1)d=a+53d+132\Rightarrow (n-1)\times12=53\times12+132$$
$$\Rightarrow n-1=53+11$$
$$\Rightarrow n=65$$
Hence, $$a_{65}$$ is $$132$$ more than $$a_{54}$$
Question 4
1 / -0

The sum of the first nineteen terms of an A.P. $$a_1,\, a_2,\, a_3$$ .......... if it is known that $$a_4\, +\, a_8\, +\, a_{12}\, +\, a_{16}\, =\, 224$$ is .......... .

A
$$1064$$

B
$$896$$

C
$$532$$

D
$$448$$

Solution

We know that $$n^{th}$$ term of an AP is given as, $$a_n = a+(n-1)d$$

And sum of first $$n$$ terms, $$S_n = \dfrac{n}{2} (2a+(n-1)d)$$, where $$a$$ and $$d$$ are the first term amd common difference of an AP.
Since, $$ { a }_{ 4 }+{ a }_{ 8 }+{ a }_{ 12 }+{ a }_{ 16 }=224$$

$$\therefore a+3d+a+7d+a+11d+a+15d=224\\ \Rightarrow 4a+36d=224\\\Rightarrow a+9d=56$$
Now
$$\displaystyle { S }_{ 19 }=\frac { 19 }{ 2 } \left[ 2a+\left( 19-1 \right) d \right] \\ =19\left[ a+9d \right] \\=19\left[ 56 \right] \\ =1064$$
Question 5
1 / -0

If $$9^{th}$$ and $$19^{th}$$ terms of an AP are $$35$$ and $$75$$, then its $$20^{th}$$ term is

A
$$78$$

B
$$79$$

C
$$80$$

D
$$81$$

Solution

$${ T }_{ 9 }=35\Rightarrow a+\left( 9-1 \right) d=35\Rightarrow a+8d=35$$
$$ { T }_{ 19 }=75\Rightarrow a+\left( 19-1 \right) d=75\Rightarrow a+18d=75$$

Solving these two we get
$$d=4,a=3$$

Therefore,
$${ T }_{ 20 }=a+\left( 20-1 \right) d=3+19\times 4=79$$
Question 6
1 / -0

The sum of all integers between $$81$$ and $$719$$ which are divisible by $$5$$ is

A
$$51800$$

B
$$50800$$

C
$$52800$$

D
None of these

Solution

All integers between $$81$$ and $$719$$ which are divisible by $$5$$ form an AP with common difference $$5$$.

First term $$a=85$$ , last term $$l=715$$ and common difference is $$d=5$$

Then, apply the formula for $$n^{th}$$ term of an AP

$$715=85+\left( n-1 \right) 5$$
$$126=n-1$$
$$\Rightarrow n=127$$

To get the required sum, apply the formula for sum of $$n$$ terms of an AP

$$\displaystyle { S }_{ n }=\dfrac { n }{ 2 } \left( a+l \right) \\=\dfrac { 127 }{ 2 } \left( 85+715 \right)\\ =50800$$
Question 7
1 / -0

The value of $$1 + 3 + 5 + 7 + 9 + ............. + 25$$ is:

A
$$196$$

B
$$625$$

C
$$225$$

D
$$169$$

Solution

We take the first term $$a=1$$, last term $$l=25$$, common difference $$d=2$$ and the number of terms $$=n$$.
Then, since it is an A.P. series,
$$l=a+(n-1)d$$
$$ n=\dfrac { l-a }{ d } +1\\ =\dfrac { 25-1 }{ 2 } +1\\$$
$$=\dfrac{25-1+2}{2}$$

$$=\dfrac{25+1}{2}$$

$$=\dfrac{26}{2}$$
$$=13$$
Let $${ S }_{ n }$$ be the sum of $$n=13$$ terms
$${ \therefore \quad S }_{ n }=\dfrac { n }{ 2 } \times \left( a+l \right) \\ =\dfrac { 13 }{ 2 } \times \left( 1+25 \right) \\ =13\times 13=169.$$
So the required sum $$=169$$.
Question 8
1 / -0

The sum of $$3^{rd}$$ and $$15^{th}$$ elements of an arithmetic progression is equal to the sum of $$6^{th},\, 11^{th}\, and\, 13^{th}$$ elements of the same progression. Then which element of the series should necessarily be equal to zero?

A
$$1st$$

B
$$9th$$

C
$$12th$$

D
None of the above

Solution

Let the first term of AP be $$a$$ and difference be $$d$$
We know that $$n$$th term, $$a_n = a+(n-1)d$$, where $$a$$ & $$d$$ are the first term amd common difference of an AP.
$$\therefore$$ Then third term will be $$=a+2d$$
$$15^{th}$$ will be$$=a+14d$$
$$6^{th}$$ will be$$ =a+5d$$
$$11^{th}$$ will be$$ =a+10d$$
$$13^{th}$$ will be$$=a+12d$$
Then the equation will be

$$a+2d+a+14d = a+5d+a+10d+a+12d$$
$$2a+16d = 3a+27d$$
$$a+11d=0$$

We understand $$a+11d$$ will be the $$12^{th}$$ term of arithmetic progression.
So, the correct answer is $$12$$.
Question 9
1 / -0

Ramkali saved $$Rs. 5$$ in the first week of a year and then increased her weekly savings by $$Rs. 1.75.$$ If in the nth week, her weekly savings become $$Rs. 20.75,$$ find $$n.$$

A
$$n=10$$

B
$$n=12$$

C
$$n=16$$

D
$$n=20$$

Solution

Given that
$$ a =5$$
$$d = 1.75$$
$$a_{n} =20.75$$
$$a_{n}=a+\left(n-1\right)d$$
$$20.75=5+\left(n-1\right)1.75$$
$$15.75=\left(n-1\right)1.75$$

$$\left(n-1\right)=\displaystyle \frac{15.75}{1.75}=\frac{1575}{175}=\frac{63}{7}=9$$
$$\therefore n-1 = 9$$
$$\therefore n=10$$
Question 10
1 / -0

The sum of first $$n$$ terms of the $$A.P$$. $$\sqrt{2}\, +\, \sqrt{8}\, +\, \sqrt{18}\, +\, \sqrt{32}\, +\, ........$$ is ......

A
$$\displaystyle \frac{n(n\, +\, 1)}{2}$$

B
$$2n (n + 1)$$

C
$$\displaystyle \frac{n(n\, +\, 1)}{\sqrt{2}}$$

D
$$1$$

Solution

The series can be rewritten as $$\sqrt{2} + 2 \sqrt{2} + 3\sqrt{2} + 4 \sqrt{2} + ..... $$
So the nth sum $$S_{n} = \sqrt{2} \times \left(1 + 2 + 3 + ..... + n \right)$$
$$= \displaystyle \frac{\sqrt{2}\times n\times \left(n+1\right)}{2} $$
$$=\displaystyle \frac{n\left(n+1\right)}{\sqrt{2}}$$

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Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 42

Result

Arithmetic Progressions Test - 42

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SHARING IS CARING

Question 1

$$131$$

$$132$$

$$133$$

$$134$$

Solution

Question 2

$$10$$

$$-10$$

$$5$$

$$-5$$

Solution

Question 3

t$$_{65}$$

t$$_{64}$$

t$$_{60}$$

None of these

Solution

Question 4

$$1064$$

$$896$$

$$532$$

$$448$$

Solution

Question 5

$$78$$

$$79$$

$$80$$

$$81$$

Solution

Question 6

$$51800$$

$$50800$$

$$52800$$

None of these

Solution

Question 7

$$196$$

$$625$$

$$225$$

$$169$$

Solution

Question 8

$$1st$$

$$9th$$

$$12th$$

None of the above

Solution

Question 9

$$n=10$$

$$n=12$$

$$n=16$$

$$n=20$$

Solution

Question 10

$$\displaystyle \frac{n(n\, +\, 1)}{2}$$

$$2n (n + 1)$$

$$\displaystyle \frac{n(n\, +\, 1)}{\sqrt{2}}$$

$$1$$

Solution

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