Chapter 5: Arithmetic Progressions

Ans: Let ${{n}^{th}}$ term of A.P. be \[132\] more than its \[{{54}^{th}}\] term i.e.,

${{a}_{n}}={{a}_{54}}+132$ …..(1)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     ….. (2)

For \[{{54}^{th}}\] term substitute $n=54$ in (2) i.e., ${{a}_{54}}=a+53d$  ….. (3)

Therefore, from (1), (2) and (3) we get, 

$a+\left( n-1 \right)d=a+53d+132$

$\Rightarrow \left( n-1 \right)d-53d=132$

$\therefore d=\dfrac{132}{n-54}$  ….. (4)

Now, given A.P. \[3,15,27,39,...\] 

Common difference $d=15-3=12$  ….. (5)

Hence, from (4) and (5) we get $12=\dfrac{132}{n-54}$

$\Rightarrow n-54=11$ 

$\therefore n=65$ 

Therefore, ${{65}^{th}}$ term of the given A.P. will be \[132\] more than its \[{{54}^{th}}\] term.

Ans: Let $2$ A.P.’s be 

$a,a+d,a+2d,a+3d,....$   …..(1)

$b,b+d,b+2d,b+3d,....$    …..(2)

(Since common difference is same)

Given that the difference between their \[{{100}^{th}}\] term is \[100\] i.e., 

${{a}_{100}}-{{b}_{100}}=100$ …..(3)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     ….. (4)

Therefore, from (3) and (4) we get, 

$a+\left( 100-1 \right)d-\left( b+\left( 100-1 \right)d \right)=100$

$\Rightarrow a-b=100$  ….. (5)

Similarly, the difference between their \[{{1000}^{th}}\] terms is, 

${{a}_{1000}}-{{b}_{1000}}=\left[ a+\left( 1000-a \right)d \right]-\left[ b+\left( 1000-a \right)d \right]$

$\Rightarrow {{a}_{1000}}-{{b}_{1000}}=a-b$

$\therefore {{a}_{1000}}-{{b}_{1000}}=100$ 

Therefore, the difference between their \[{{1000}^{th}}\] terms is $100$.

Ans: First three-digit number that is divisible by $7$ is \[105\] then the next number will be \[105+7=112\]. 

Therefore, the series becomes \[105,112,119,....\] 

This is an A.P. having first term as \[105\] and common difference as \[7\].

Now, the largest $3$ digit number is $999$.

Leu us divide it by \[7\] , to get the remainder.

$999=142\times 7+5$ 

Therefore, \[999-5=994\] is the maximum possible three-digit number that is divisible by $7$.

Also, this will be the last term of the A.P. series.

Hence the final series is as follows: \[105,112,119,...,994\] 

Let 994 be the \[{{n}^{th}}\] term of this A.P.

Then, \[{{a}_{n}}=105+7\left( n-1 \right)\] 

\[\Rightarrow 994=105+7\left( n-1 \right)\]

\[\Rightarrow 889=7\left( n-1 \right)\]

\[\Rightarrow 127=\left( n-1 \right)\]

$\therefore n=128$ 

Therefore, $128$ three-digit numbers are divisible by $7$.

Ans: First number that is divisible by $4$ and lie between $10$ and $250$ is $12$. The next number will be \[12+4=16\]. 

Therefore, the series becomes \[12,16,20,....\] 

This is an A.P. having first term as \[12\] and common difference as \[4\].

Now, the largest number in range is $250$.

Leu us divide it by \[4\] to get the remainder.

$250=62\times 4+2$ 

Therefore, \[250-2=248\] is the last term of the A.P. series.

Hence the final series is as follows: \[12,16,20,....,248\] 

Let $248$ be the \[{{n}^{th}}\] term of this A.P.

Then, \[{{a}_{n}}=12+4\left( n-1 \right)\] 

\[\Rightarrow 248=12+4\left( n-1 \right)\]

\[\Rightarrow 236=4\left( n-1 \right)\]

\[\Rightarrow 59=\left( n-1 \right)\]

$\therefore n=60$ 

Therefore, $60$ multiples of $4$ lie between $10$ and $250$.

 Ans: Given $2$ A.P.’s are

\[63,65,67,....\]   …..(1)

Its first term is $63$ and common difference is $65-63=2$ 

\[3,10,17,....\]    …..(2)

Its first term is $3$ and common difference is $10-3=7$ 

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     ….. (3)

Therefore, from (1) and (3) we get the ${{n}^{th}}$ term of the first A.P. is

 ${{a}_{n}}=63+2\left( n-1 \right)$

$\Rightarrow {{a}_{n}}=61+2n$  ….. (4)

And from (2) and (3) we get the ${{n}^{th}}$ term of the second A.P. is

 ${{b}_{n}}=3+7\left( n-1 \right)$

$\Rightarrow {{b}_{n}}=-4+7n$  ….. (5)

If the ${{n}^{th}}$ terms of two APs \[63,65,67,....\] and \[3,10,17,....\] are equal the from (4) and (5),

\[{{a}_{n}}={{b}_{n}}\]

$\Rightarrow 61+2n=-4+7n$

$\Rightarrow 65=5n$

$\therefore n=13$ 

Therefore, the ${{13}^{th}}$ term of both the A.P.’s are equal.

Ans: Given the ${{7}^{th}}$ term of A.P. is \[12\] more than its \[{{5}^{th}}\] term i.e.,

${{a}_{7}}={{a}_{5}}+12$ …..(1)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     ….. (2)

For \[{{5}^{th}}\] term substitute $n=5$ in (2) i.e., ${{a}_{5}}=a+4d$  ….. (3)

For \[{{7}^{th}}\] term substitute $n=7$ in (2) i.e., ${{a}_{7}}=a+6d$  ….. (4)

Therefore, from (1), (3) and (4) we get, 

$a+6d=a+4d+12$

$\Rightarrow 2d=12$

$\therefore d=6$   ….. (5)

Substituting (5) in (2) we get, ${{a}_{n}}=a+6\left( n-1 \right)$ ……(6)

Given the third term of the A.P. is \[16\]. Hence from (6),

$16=a+6\left( 3-1 \right)$

$\Rightarrow 16=a+12$

$\therefore a=4$   ….. (7)

Hence from (6), ${{a}_{n}}=4+6\left( n-1 \right)$

Therefore, the A.P. will be \[4,10,16,22,....\].

Ans: Given A.P. \[3,8,13,...,253\]. To find the \[{{20}^{th}}\] term from the last write the given A.P. in reverse order and then find its \[{{20}^{th}}\] term.

Required A.P. is $253,...,13,8,3$  ….. (1) 

Its first A.P. is $253$ and common difference is $8-13=-5$.  …..(2)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     ….. (3)

Hence from (2) and (3) we get, ${{a}_{n}}=253-5\left( n-1 \right)$   …..(4)

Substitute $n=20$ in (4) we get, 

${{a}_{20}}=253-5\left( 20-1 \right)$

$\Rightarrow {{a}_{20}}=158$.

Therefore, ${{20}^{th}}$ term from the last is $158$.

Ans: Given the sum of \[{{4}^{th}}\] and \[{{8}^{th}}\] terms of an A.P. is \[24\] i.e.,

${{a}_{4}}+{{a}_{8}}=24$ …..(1)

Given the sum of \[{{6}^{th}}\] and \[{{10}^{th}}\] terms is \[44\] i.e.,

${{a}_{6}}+{{a}_{10}}=44$ …..(2)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     ….. (3)

For \[{{4}^{th}}\] term substitute $n=4$ in (3) i.e., ${{a}_{4}}=a+3d$  ….. (4)

For \[{{6}^{th}}\] term substitute $n=6$ in (3) i.e., ${{a}_{6}}=a+5d$  ….. (5)

For \[{{8}^{th}}\] term substitute $n=8$ in (3) i.e., ${{a}_{8}}=a+7d$  ….. (6)

For \[{{10}^{th}}\] term substitute $n=10$ in (3) i.e., ${{a}_{10}}=a+9d$  ….. (7)

Therefore, from (1), (6) and (4) we get, 

$\left( a+3d \right)+\left( a+7d \right)=24$

$\Rightarrow 2a+10d=24$

$\Rightarrow a+5d=12$   ….. (8)

From (2), (5) and (7) we get, 

$\left( a+5d \right)+\left( a+9d \right)=44$

$\Rightarrow 2a+14d=44$

$\Rightarrow a+7d=22$   ….. (9)

Subtracting (8) from (9) we get,  

$\Rightarrow \left( a+7d \right)-\left( a+5d \right)=22-12$

$\Rightarrow 2d=10$

$\therefore d=5$ ….. (10)

Substituting this value from (10) in (9) we get,

$a+35=22$

$\therefore a=-13$ ….. (11)

Therefore from (10) and (11), the first three terms of the A.P. are $-13,-8,-3$.

Ans: Given in the first year, annual salary is Rs $5000$.  

In the second year, annual salary is Rs $5000+200=5200$.  

In the third year, annual salary is Rs $5200+200=5400$. 

This series will form an A.P. with first term $5000$ and common difference $200$.  

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$  

Therefore, In the ${{n}^{th}}$ year, annual salary is ${{a}_{n}}=5000+200\left( n-1 \right)$

$\Rightarrow {{a}_{n}}=4800+200n$ …. (1)

To find the year in which his annual income reaches Rs \[7000\], substitute ${{a}_{n}}=7000$ in (1) and find the value of $n$i.e., 

  $7000=4800+200n$

$\Rightarrow 2200+200n$

$\therefore n=11$ 

Therefore, in ${{11}^{th}}$ year i.e., in $2005$ his salary will be Rs $7000$.

Ans: Given in the first week the savings is Rs $5$.  

In the second week the savings is Rs $5+1.75=6.75$.  

In the third week the savings is Rs $6.75+1.75=8.5$. 

This series will form an A.P. with first term $5$ and common difference $1.75$.  

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$  

Therefore, In the ${{n}^{th}}$ week the savings is ${{a}_{n}}=5+1.75\left( n-1 \right)$

$\Rightarrow {{a}_{n}}=3.25+1.75n$ …. (1)

To find the week in which her savings reaches Rs \[20.75\], substitute ${{a}_{n}}=20.75$ in (1) and find the value of $n$i.e., 

  $20.75=3.25+1.75n$

$\Rightarrow 17.5=1.75n$

$\therefore n=10$ 

Therefore, in ${{10}^{th}}$ week her savings will be Rs $20.75$.