# CBSE 10th Mathematics | Arithmetic Progressions | nth Term of an Arithmetic Progression

## nth Term of an Arithmetic Progression

Let $\displaystyle {{a}_{1}},{{a}_{2}},{{a}_{3}}$…….. be an A.P., with first term as a, and common difference as d.

First term is         = a            (i)

Second term $\displaystyle ({{a}_{2}})$     = a + d            (ii)

= $\displaystyle a+(2-1)d$

Third term $\displaystyle ({{a}_{3}})$        = $\displaystyle {{a}_{2}}+d$        (iii)

= $\displaystyle a+d+d$      [from (i)]

= $\displaystyle a+2d$

or         = $\displaystyle a+(3-1)d$

Fourth term $\displaystyle {{a}_{4}}={{a}_{3}}+d$

or          = $\displaystyle a+2d+d$     [from (iii)]

= $\displaystyle a+3d$

= $\displaystyle a+(4-1)d$

\      nth term an $\displaystyle =a+(n-1)d$

Note: The nth term of the A.P. with first term a & common difference d is given by $\displaystyle {{a}_{n}}=a+(n-1)d$

$\displaystyle {{a}_{n}}$ is also called as general term of an A.P.

If there are P terms in the A.P. then ap represents the last term which can also be denoted by l.

## Problems based on nth TERM OF AN Arithmetic Progression

### Solution:

Here      a = 7

and      d = $\displaystyle {{a}_{2}}-{{a}_{1}}$

= 4 7 = 3

n = 18

$\displaystyle {{a}_{n}}=a+\left( n-1 \right)d$

$\displaystyle {{a}_{18}}=7+\left( 18-1 \right)\times -3$

= 7 + 17 ´ 3

= 7 51

= 44

$\displaystyle {{a}_{n}}=a+\left( n-1 \right)d$

= 7 + ( 1) (3)

= 7 3n + 3

= 10 3n

### Solution:

a = 7

$\displaystyle d={{a}_{2}}-{{a}_{1}}$

= 12 7 = 5

$\displaystyle {{a}_{n}}=87$

As     $\displaystyle {{a}_{n}}=a+\left( n-1 \right)d$

87 = 7 + ( 1) ´ 5

$\displaystyle 87-7=5n-5$

$\displaystyle 80+5=5n$

$\displaystyle \frac{85}{5}=n$

17 = n

\     17th term of give A.P. is 87

### Solution:

a = 7

$\displaystyle d={{a}_{2}}-{{a}_{1}}$

= 13 7 = 6

$\displaystyle {{a}_{n}}=a+\left( n-1 \right)d$

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

205 7 = 6n 6

198 + 6 = 6n

$\displaystyle \frac{204}{6}=n$ or 34 = n

\     Given A.P. has 34 terms

### Solution:

a = 11

$\displaystyle d={{a}_{2}}-{{a}_{1}}$

= 8 11 = -3

let $\displaystyle {{a}_{n}}=-150$                [Assume]

$\displaystyle -150-11=-3n+3$

$\displaystyle -161-3=-3n$

$\displaystyle -164=-3n$

$\displaystyle \frac{164}{3}=n$

As number of term can’t be in fraction, 150 is not a term of the given A.P.

### Solution:

(i)    Here a = 5

& $\displaystyle {{a}_{4}}=9\frac{1}{2}$ or $\displaystyle \frac{19}{2}$

Now,      $\displaystyle {{a}_{4}}=a+3d$

$\displaystyle \frac{19}{2}=5+3d$

$\displaystyle \frac{19}{2}-5=3d$

$\displaystyle \frac{9}{2}=3d$

$\displaystyle \frac{9}{6}=d$

\     $\displaystyle d=\frac{3}{2}$

\     $\displaystyle {{a}_{2}}=a+d=5+\frac{3}{2}=\frac{13}{2}$

$\displaystyle {{a}_{3}}={{a}_{2}}+d=\frac{13}{2}+\frac{3}{2}=\frac{16}{2}=8$

Hence

(ii)     Here $\displaystyle a=-4$

$\displaystyle {{a}_{6}}=6$

As    $\displaystyle {{a}_{6}}=a+5d$

6 = 4 + 5d

$\displaystyle \frac{10}{5}=d$

d = 2

\     $\displaystyle {{a}_{2}}=a+d=-4+2=-2$

$\displaystyle {{a}_{3}}={{a}_{2}}+d=-2+2=0$

$\displaystyle {{a}_{4}}={{a}_{3}}+d=0+2=2$

$\displaystyle {{a}_{5}}={{a}_{4}}+d=2+2=4$

### Solution:

First sequence is 63, 65, 67, ……..

a = 63

d1 = 65 63 = 2

$\displaystyle {{a}_{n}}=63+\left( n-1 \right)2$

= 63 + 2n2

= 61 + 2n

Second sequence is 3, 10, 17, ………..

\ $\displaystyle b=3,\,\,\,{{d}_{2}}=7$

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

=     $\displaystyle 3+7n-7$

= 7n4

According to question

61 + 2n = 7n 4

61 + 4 = 7n 2n

65 = 5n

$\displaystyle \frac{65}{5}=n$

\n = 13

### Solution:

Let the common difference of two A.P.s be d

Then, their 100th terms will be

$\displaystyle {{a}_{100}}={{a}_{1}}+99d;\,\,\,\,\,{{b}_{100}}={{b}_{1}}+99d$

According to question     $\displaystyle {{a}_{100}}-{{b}_{100}}=100$

i.e.            …(i)

Now, difference between their 1000th terms

$\displaystyle ={{a}_{1000}}-{{b}_{1000}}=\left( {{a}_{1}}+999d \right)-\left( {{b}_{1}}+999d \right)$

$\displaystyle ={{a}_{1}}-{{b}_{1}}=100$        [By equation (i)]

### Solution:

The give A.P. is 3, 10, 17, ……

Here, a (first term)     = 3

d (common difference)= 10 3 = 7

$\displaystyle {{a}_{n}}=a+\left( n-1 \right)d$

$\displaystyle {{a}_{13}}=3+(13-1)7$

= 3 + 12 ´ 7 = 3 + 84 = 87

Let nth term be 84 more than the 13th term of the given A.P.

So, we get     $\displaystyle {{a}_{n}}=84+{{a}_{13}}$

$\displaystyle a+\left( n-1 \right)d=84+a+12d$

$\displaystyle d\left( n-1-12 \right)=84$

$\displaystyle 7\left( n-13 \right)=84$

$\displaystyle n-13=\frac{84}{7}=12$

$\displaystyle n=12+13=25$

\ The 25th term of the given A.P. will be 84 more than its 13th term.

### Solution:

Let a be the 1st term and d the common difference.

Here    $\displaystyle {{a}_{11}}=a+10d=38$                …(i)

$\displaystyle {{a}_{16}}=a+15d=73$                …(ii)

Subtracting (ii) and (i), we get

$\displaystyle a+10d-a-15d=38-73$

or

Putting d = 7 in (i), we get

$\displaystyle a+10\times 7=38$

$\displaystyle a=38-70=-32$

\     $\displaystyle {{a}_{31}}=a+30d=-32+30\times 7=-32+210=178$

Hence, 31st term is 178.

### Solution:

Three digit numbers which are divisible by 7 are 105, 112, 119, …, 994.

Here,      $\displaystyle a=105,\,\,\,\,\,d=7,\,\,\,\,\,{{a}_{n}}=994$

\     $\displaystyle a+(n-1)d=994$

$\displaystyle 105+(n-1)7=994$

Þ     $\displaystyle 7(n-1)=994-105=889$

Þ         $\displaystyle n-1=\frac{889}{7}$

Þ         1 = 127

n = 127 + 1 = 128.

\    There are 128 three digit numbers which are divisible by 7

### Solution:

We know that the formula to calculate simple interest is given by

Simple Interest = $\displaystyle \frac{P\times R\times T}{100}$

So, the interest at the end of the 1st year     = Rs $\displaystyle \frac{1000\times 8\times 1}{100}$ = Rs 80

The interest at the end of the 2nd year     = Rs $\displaystyle \frac{1000\times 8\times 2}{100}$ = Rs 160

The interest at the end of the 3rd year     = Rs $\displaystyle \frac{1000\times 8\times 3}{100}$ = Rs 240

Similarly, we can obtain the interest at the end of the 4th year, 5th year, and so on.

So, the interest (in Rs) at the end of the 1st, 2nd, 3rd, ….. years, respectively are 80, 160, 240, …

It is an AP as the difference between the consecutive terns in the list is 80, i.e., d = 80. Also, a = 80.

So, to find the interest at the end of 30 years, we shall find $\displaystyle {{a}_{30}}$.

Now, $\displaystyle {{a}_{30}}=a+(30-1)d=80+29\times 80=2400$

So, the interest at the end of 30 years will be Rs 2400.

## TO FIND nth TERM FROM THE END OF AN A.P.

Consider the following A.P. $\displaystyle a,a+d,\,a+2d,....\left( l-2d \right),\left( l-d \right)$, $\displaystyle l$

where l is the last term

last term          $\displaystyle l$ = $\displaystyle l$ (1 1)d

2nd last term     $\displaystyle l$
d = $\displaystyle l$ (2 1) d

3rd last term      $\displaystyle l$ 2d = $\displaystyle l$ (3 1)d

……………………………

……………………………

Note: nth term from the end = $\displaystyle l$  (n 1) d

### Solution:

1st method

$\displaystyle l=-40,d=14-17=-3$

Using $\displaystyle l-(n-1)d$

5th term from the end will be

$\displaystyle =-40-(5-1)\times -3$

$\displaystyle =-40-4\times -3$

$\displaystyle =-40+12$

= 28

2nd method

Sequence can be written as 40, 37, ….11, 14, 17

\     $\displaystyle a=-40$

$\displaystyle d=-37-(-40)$

=37 + 40

= 3

n = 5

Using $\displaystyle {{a}_{n}}=a+(n-1)d$

=$\displaystyle -40+(5-1)\times 3$

$\displaystyle =-40+4\times 3$

= 40 + 12

= 28

### Solution:

Using 1st condition

\     $\displaystyle {{a}_{4}}+{{a}_{8}}=24$

$\displaystyle a+3d+a+7d=24$

$\displaystyle 2a+10d=24$

or     $\displaystyle a+5d=12$            …(i)

Using 2nd condition

$\displaystyle {{a}_{6}}+{{a}_{10}}=44$

or    $\displaystyle a+5d+a+9d=44$

$\displaystyle 2a+14d=44$

or    $\displaystyle a+7d=22$            …(ii)

Subtracting equation (i) from (ii) we have

$\displaystyle 2d=10$

or     d = 5

Putting value of d in equations (i)

$\displaystyle a+5\times 5=12$

$\displaystyle a=12-25$

$\displaystyle a=-13$

\ $\displaystyle {{a}_{2}}=a+d$

$\displaystyle =-13+5=-8$

$\displaystyle {{a}_{3}}=a+2d$

$\displaystyle =-13+2\times 5$

$\displaystyle =-13+10=-3$

### Solution:

Let a be the first term and d be the common difference of the given A.P.

Then,     pth term = q    Þ    $\displaystyle a+\left( p-1 \right)d=q$    …(i)

qth term = p    Þ    $\displaystyle a+\left( q-1 \right)d=p$    …(ii)

Subtracting equation (ii) from equation (i), we get

…(iii)

Putting $\displaystyle d=-1$ in equation (i), we get

$\displaystyle a+1-p=q$

\    nth term $\displaystyle =a+\left( n-1 \right)d$

$\displaystyle =\left( p+q-1 \right)+\left( n-1 \right)\times (-1)$     [from equation (iii)]

$\displaystyle =\left( p+q-n \right)$

### Solution:

Let a be the first term and d be the common difference of the given A.P.

According to question, m times the mth term = n times the nth term

Þ    $\displaystyle m{{a}_{m}}=n{{a}_{n}}$

Þ    $\displaystyle m\left\{ a+\left( m-1 \right)d \right\}=n\left\{ a+\left( n-1 \right)d \right\}$

Þ    $\displaystyle m\left\{ a+\left( m-1 \right)d \right\}-n\left\{ a+\left( n-1 \right)d \right\}=0$

Þ    $\displaystyle a\left( m-n \right)+\left\{ m\left( m-1 \right)-n\left( n-1 \right) \right\}d=0$

Þ    $\displaystyle a\left( m-n \right)+\left( {{m}^{2}}-{{n}^{2}}-m+n \right)d=0$

Þ    $\displaystyle a\left( m-n \right)+\left[ \left( m-n \right)\left( m+n \right)-\left( m-n \right) \right]d=0$

Þ    $\displaystyle a\left( m-n \right)+\left[ \left( m-n \right)\left( m+n-1 \right)d \right]=0$

Þ    $\displaystyle \left( m-n \right)\left[ a+\left( m+n-1 \right)d \right]=0$

Þ    $\displaystyle a+\left( m+n-1 \right)d=0$        [$\displaystyle \because m\ne n$]

Þ    $\displaystyle {{a}_{m+n}}=0$

\     $\displaystyle \left( m+n \right)$th term of A.P. is zero

## CONDITION FOR TERMS TO BE IN A.P.

If three numbers a, b, c, in order are in A.P. Then,

$\displaystyle b-a=$ common difference = $\displaystyle c-b$

Þ    $\displaystyle b-a=c-b$

Þ    $\displaystyle 2b=a+c$

Note: a, b, c are in A.P. iff $\displaystyle 2b=a+c$

### Solution:

Since, $\displaystyle 2x,x+10,3x+2$ are in A.P.

\     $\displaystyle 2\left( x+10 \right)=2x+\left( 3x+2 \right)$

Þ     $\displaystyle 2x+20=5x+2$

Þ     $\displaystyle 3x=18$

Þ     x = 6

### Solution:

Let D be the common difference of the given A.P. Then,

$\displaystyle b=a+D,c=a+2D,d=a+3D$ and $\displaystyle e=a+4D$

\    $\displaystyle a-4b+6c-4d+e=a-4(a+D)+6(a+2D)-4(a+3D)+(a+4D)$

Þ    $\displaystyle a-4b+6c-4d+e=a-4a-4D+6a+12D-4a-12D+a+4D$

Þ    $\displaystyle a-4b+6c-4d+e=a-4a+6a-4a+a-4D+12D-12D+4D$

Þ    $\displaystyle a-4b+6c-4d+e=0$

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