Chapter 5: Arithmetic Progressions

Ans: 

i. Given, the first Term, $a=7$  ….. (1)

Given, the common Difference, \[d=3\] …..(2)

Given, the number of Terms, \[n=8\]  …..(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)

Substituting the values from (1), (2) and (3) in (4) we get,

${{a}_{n}}=7+\left( 8-1 \right)3$

$\Rightarrow {{a}_{n}}=7+21$

$\therefore {{a}_{n}}=28$

ii. Given, the first Term, $a=-18$  ….. (1)

Given, the ${{n}^{th}}$ term, \[{{a}_{n}}=0\] …..(2)

Given, the number of Terms, \[n=10\]  …..(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)

Substituting the values from (1), (2) and (3) in (4) we get,

$0=-18+\left( 10-1 \right)d$

$\Rightarrow 18=9d$

$\therefore d=2$

iii. Given, the ${{n}^{th}}$ term, \[{{a}_{n}}=-5\] ….. (1)

Given, the common Difference, \[d=-3\] …..(2)

Given, the number of Terms, \[n=18\]  …..(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)

Substituting the values from (1), (2) and (3) in (4) we get,

$-5=a+\left( 18-1 \right)\left( -3 \right)$

$\Rightarrow -5=a-51$

$\therefore a = 46$

iv. Given, the first Term, $a=-18.9$  ….. (1)

Given, the common Difference, \[d=2.5\] …..(2)

Given, the ${{n}^{th}}$ term, \[{{a}_{n}}=3.6\] …..(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)

Substituting the values from (1), (2) and (3) in (4) we get,

$3.6=-18.9+\left( n-1 \right)\left( 2.5 \right)$

$\Rightarrow 22.5=\left( n-1 \right)\left( 2.5 \right)$

$\Rightarrow 9=\left( n-1 \right)$

$\therefore n=10$

v. Given, the first Term, $a=3.5$  ….. (1)

Given, the common Difference, \[d=0\] …..(2)

Given, the number of Terms, \[n=105\]  …..(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)

Substituting the values from (1), (2) and (3) in (4) we get,

${{a}_{n}}=3.5+\left( 105-1 \right)\left( 0 \right)$

$\therefore {{a}_{n}}=3.5$ 

Ans: Option C. $-77$ 

Given, the first Term, $a=10$  ….. (1)

Given, the common Difference, \[d=7-10=-3\] …..(2)

Given, the number of Terms, \[n=30\]  …..(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)

Substituting the values from (1), (2) and (3) in (4) we get, ${{a}_{30}}=10+\left( 30-1 \right)\left( -3 \right)$

$\Rightarrow {{a}_{30}}=10-87$

$\therefore {{a}_{30}}=-77$

ii. \[{{11}^{th}}\] term of the A.P \[-3,-\dfrac{1}{2},2,...,\] is

1. \[28\]

2. \[22\]  

3. \[38\] 

4. \[48\dfrac{1}{2}\] 

Ans: Option II, $22$ 

Given, the first Term, $a=-3$  ….. (1)

Given, the common Difference, \[d=-\dfrac{1}{2}-\left( -3 \right)=\dfrac{5}{2}\] …..(2)

Given, the number of Terms, \[n=11\]  …..(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)

Substituting the values from (1), (2) and (3) in (4) we get, ${{a}_{n}}=-3+\dfrac{5}{2}\left( 11-1 \right)$

$\Rightarrow {{a}_{n}}=-3+25$

$\therefore {{a}_{n}}=22$ 

Ans: Given, first term $a=2$ …..(1)

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

Substituting the values from (1) we get, ${{a}_{n}}=2+\left( n-1 \right)d$  …..(2)

Given, third term ${{a}_{3}}=26$. From (2) we get,

$26=2+\left( 3-1 \right)d$

$\Rightarrow 26=2+2d$ 

$\therefore d=12$  ….(3)

From (1), (2) and (3) we get for $n=2$ 

${{a}_{2}}=2+\left( 2-1 \right)\left( 12 \right)$

$\therefore {{a}_{2}}=14$

$\therefore $ The sequence is \[2,14,26\].

Ans: Given, second term ${{a}_{2}}=13$ …..(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)

Substituting the values from (1) for $n=2$ we get, $13=a+d$  …..(3)

Given, fourth term ${{a}_{4}}=3$. From (2) we get, 

$3=a+3d$  …..(4)

Solving (3) and (4) by subtracting (3) from (4) we get,

$3-13=\left( a+3d \right)-\left( a+d \right)$

$\Rightarrow -10=2d$ 

$\therefore d=-5$  ….(5)

From (3) and (5) we get 

$13=a-5$

$\Rightarrow a=18$ ……(6)

Substituting the values from (5) and (6) in (2) we get,

${{a}_{n}}=18-5\left( n-1 \right)$   …..(7)

First term, $a=18$ and third term ${{a}_{3}}=8$

$\therefore $ The sequence is \[18,13,8,3\].

Ans: Given, first term $a=5$ …..(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)

Substituting the values from (1) in (2) we get, ${{a}_{n}}=5+\left( n-1 \right)d$  …..(3)

Given, fourth term ${{a}_{4}}=9\dfrac{1}{2}$. From (3) we get,

$9\dfrac{1}{2}=5+\left( 4-1 \right)d$ 

$\Rightarrow 9\dfrac{1}{2}=5+3d$ 

$\therefore d=\dfrac{3}{2}$  ….(4)

From (3) and (4) we get 

${{a}_{n}}=5+\dfrac{3}{2}\left( n-1 \right)$ ……(5)

Second term, ${{a}_{2}}=\dfrac{13}{2}$ and third term ${{a}_{3}}=8$

$\therefore $ The sequence is $5,\dfrac{13}{2},8,9\dfrac{1}{2}$.

Ans: Given, first term $a=-4$ …..(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)

Substituting the values from (1) in (2) we get, ${{a}_{n}}=-4+\left( n-1 \right)d$  …..(3)

Given, sixth term ${{a}_{6}}=6$. From (3) we get, 

$6=-4+\left( 6-1 \right)d$ 

$\Rightarrow 6=-4+5d$ 

$\therefore d=2$  ….(4)

From (3) and (4) we get 

${{a}_{n}}=-4+2\left( n-1 \right)$ ……(5)

Second term ${{a}_{2}}=-2$, third term ${{a}_{3}}=0$, fourth term ${{a}_{4}}=2$ and fifth term ${{a}_{5}}=4$.

$\therefore $ The sequence is $-4,-2,0,2,4,6$

Ans: Given, second term ${{a}_{2}}=38$ …..(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)

Substituting the values from (1) for $n=2$ we get, $38=a+d$  …..(3)

Given, sixth term ${{a}_{6}}=-22$. From (2) we get,

$-22=a+5d$  …..(4)

Solving (3) and (4) by subtracting (3) from (4) we get,

$-22-38=\left( a+5d \right)-\left( a+d \right)$

$\Rightarrow -60=4d$ 

$\therefore d=-15$  ….(5)

From (3) and (5) we get 

$38=a-15$

$\Rightarrow a=53$ ……(6)

Substituting the values from (5) and (6) in (2) we get,
${{a}_{n}}=53-15\left( n-1 \right)$   …..(7)

The first term, $a=53$, second term ${{a}_{3}}=23$, third term ${{a}_{3}}=8$ and fourth term ${{a}_{4}}=-7$

$\therefore $ The sequence is \[53,38,23,8,-7,-22\].

Ans: Given, the first Term, $a=3$  ….. (1)

Given, the common Difference, \[d=8-3=5\] …..(2)

Given, the ${{n}^{th}}$ term, \[{{a}_{n}}=78\] …..(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)

Substituting the values from (1), (2) and (3) in (4) we get,

$78=3+5\left( n-1 \right)$

$\Rightarrow 75=5\left( n-1 \right)$

$\Rightarrow 15=\left( n-1 \right)$

$\therefore n=16$ 

Therefore, \[{{16}^{th}}\] term of this A.P. is \[78\].

Ans: Given, the first Term, $a=7$  ….. (1)

Given, the common Difference, \[d=13-7=6\] …..(2)

Given, the ${{n}^{th}}$ term, \[{{a}_{n}}=205\] …..(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)

Substituting the values from (1), (2) and (3) in (4) we get,

$205=7+6\left( n-1 \right)$

$\Rightarrow 198=6\left( n-1 \right)$

$\Rightarrow 33=\left( n-1 \right)$

$\therefore n=34$ 

Therefore, given A.P. series has \[34\] terms.

Ans: Given, the first Term, $a=18$  ….. (1)

Given, the common Difference, \[d=15\dfrac{1}{2}-18=-\dfrac{5}{2}\] …..(2)

Given, the ${{n}^{th}}$ term, \[{{a}_{n}}=-47\] …..(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)

Substituting the values from (1), (2) and (3) in (4) we get,

$-47=18-\dfrac{5}{2}\left( n-1 \right)$

$\Rightarrow -65=-\dfrac{5}{2}\left( n-1 \right)$

$\Rightarrow 26=\left( n-1 \right)$

$\therefore n=27$ 

Therefore, given A.P. series has \[27\] terms.

Ans: Given, the first Term, $a=11$  ….. (1)

Given, the common Difference, \[d=8-11=-3\] …..(2)

Given, the ${{n}^{th}}$ term, \[{{a}_{n}}=-150\] …..(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)

Substituting the values from (1), (2) and (3) in (4) we get,

$-150=11-3\left( n-1 \right)$

$\Rightarrow -161=-3\left( n-1 \right)$

$\Rightarrow \dfrac{161}{3}=\left( n-1 \right)$

$\therefore n=\dfrac{164}{3}$ 

Since $n$ is nor a natural number. Therefore, $-150$ is not a term of the given A.P. series.

Ans: Given, the \[{{11}^{th}}\] Term, ${{a}_{11}}=38$  ….. (1)

Given, the \[{{16}^{th}}\] Term, ${{a}_{16}}=73$  …..(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)

Substituting the values from (1) in (3) we get,

$38=a+\left( 11-1 \right)d$

$\Rightarrow 38=a+10d$      …..(4)

Substituting the values from (2) in (3) we get,

$73=a+\left( 16-1 \right)d$

$\Rightarrow 73=a+15d$      …..(5)

Solving equations (4) and (5) by subtracting (4) from (5) we get,

$\Rightarrow 73-38=\left( a+15d \right)-\left( a+10d \right)$

$\Rightarrow 5=35d$

$\therefore d=7$   …..(6)

Substituting value from (6) in (4) we get,

$\Rightarrow 38=a+70$

$\therefore a=-32$   …..(7)

Again, substituting the values from (6) and (7) in (3) we get,

${{a}_{n}}=-32+7\left( n-1 \right)$  …..(8)

To find the ${{31}^{st}}$ term substitute $n=31$ in (8) we get, 

${{a}_{31}}=-32+7\left( 31-1 \right)$

$\Rightarrow {{a}_{31}}=-32+210$

$\therefore {{a}_{31}}=178$

Therefore, the \[{{31}^{st}}\] term of an A.P. whose \[{{11}^{th}}\] term is \[38\] and the \[{{16}^{th}}\] term is \[73\] is $178$. 

Ans: Given, the \[{{3}^{rd}}\] Term, ${{a}_{3}}=12$  ….. (1)

Given, the \[{{50}^{th}}\] Term, ${{a}_{50}}=106$  …..(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)

Substituting the values from (1) in (3) we get,

$12=a+\left( 3-1 \right)d$

$\Rightarrow 12=a+2d$      …..(4)

Substituting the values from (2) in (3) we get,

$106=a+\left( 50-1 \right)d$

$\Rightarrow 106=a+49d$      …..(5)

Solving equations (4) and (5) by subtracting (4) from (5) we get,

$\Rightarrow 106-12=\left( a+49d \right)-\left( a+2d \right)$

$\Rightarrow 94=47d$

$\therefore d=2$   …..(6)

Substituting value from (6) in (4) we get,

$\Rightarrow 12=a+4$

$\therefore a=8$   …..(7)

Again, substituting the values from (6) and (7) in (3) we get,

${{a}_{n}}=8+2\left( n-1 \right)$  …..(8)

To find the ${{29}^{th}}$ term substitute $n=29$ in (8) we get, 

${{a}_{29}}=8+2\left( 29-1 \right)$

$\Rightarrow {{a}_{29}}=8+56$

$\therefore {{a}_{29}}=64$

Therefore, the ${{29}^{th}}$ term of the A.P. is $64$. 

Ans: Given, the \[{{3}^{rd}}\] Term, ${{a}_{3}}=4$  ….. (1)

Given, the \[{{9}^{th}}\] Term, ${{a}_{9}}=-8$  …..(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)

Substituting the values from (1) in (3) we get,

$4=a+\left( 3-1 \right)d$

$\Rightarrow 4=a+2d$      …..(4)

Substituting the values from (2) in (3) we get,

$-8=a+\left( 9-1 \right)d$

$\Rightarrow -8=a+8d$      …..(5)

Solving equations (4) and (5) by subtracting (4) from (5) we get,

$\Rightarrow -8-4=\left( a+8d \right)-\left( a+2d \right)$

$\Rightarrow -12=6d$

$\therefore d=-2$   …..(6)

Substituting value from (6) in (4) we get,

$\Rightarrow 4=a-4$

$\therefore a=8$   …..(7)

Again, substituting the values from (6) and (7) in (3) we get,

${{a}_{n}}=8-2\left( n-1 \right)$  …..(8)

T find the term which is zero, substitute ${{a}_{n}}=0$ in (8)

$0=8-2\left( n-1 \right)$

$\Rightarrow 8=2\left( n-1 \right)$

$\Rightarrow 4=\left( n-1 \right)$

$\therefore n=5$ 

Therefore, given A.P. series has \[{{5}^{th}}\] term as zero.

Ans: Given that the \[{{17}^{th}}\] term of an A.P. exceeds its \[{{10}^{th}}\] term by \[7\] i.e., 

${{a}_{17}}={{a}_{10}}+7$ …..(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 \[{{17}^{th}}\] term substitute $n=17$ in (2) i.e., ${{a}_{17}}=a+16d$  ….. (3)

For \[{{10}^{th}}\] term substitute $n=10$ in (2) i.e., ${{a}_{10}}=a+9d$  ….. (4)

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

$a+16d=a+9d+7$

$\Rightarrow 7d=7$

$\therefore d=1$ 

Therefore, the common difference is $1$.