# CBSE 10th Mathematics | Arithmetic Progressions | Sum of n terms of an Arithmetic Progression

## Sum of n terms of an Arithmetic Progression

Let a be the first term and d be the common difference of an A.P. $\displaystyle l$ is the last term where l = a + (n
1) d.

Sum of first n terms of the given A.P. is given by

$\displaystyle {{S}_{n}}=a+\left( a+d \right)+\left( a+2d \right)+....\left( l-2d \right)+\left( l-d \right)+l$…(i)

Writing in reverse order

$\displaystyle {{S}_{n}}=l+\left( l-d \right)+\left( l-2d \right)+....\left( a+2d \right)+\left( a+d \right)+a$…(ii)

Adding (i) and (ii) we get

$\displaystyle 2{{S}_{n}}=\underbrace{\left( a+l \right)+\left( a+l \right)+\left( a+l \right)+.....+\left( a+l \right)}_{n\,\,\,times}$

$\displaystyle 2{{S}_{n}}=n\left( a+l \right)$

$\displaystyle {{S}_{n}}=\frac{n}{2}\left( a+l \right)$

$\displaystyle =\frac{n}{2}\left[ a+a+\left( n-1 \right)d \right]$ [$\displaystyle \because$ l = a + (n – 1)d]

$\displaystyle {{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$

$\displaystyle {{S}_{n}}=\frac{n}{2}\left( a+{{a}_{n}} \right)$where an = a + (n – 1)d

## Solved Problems based on Sum of n terms of an Arithmetic Progression

### Solution:

(i)

$\displaystyle {{S}_{n}}=1+2+3+....+100$

Here, $\displaystyle a=1,\,\,\,d=2-1=1,\,\,\,\,n=100$

Using the formula

$\displaystyle {{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$

\ $\displaystyle {{S}_{n}}=\frac{100}{2}\left[ 2\times 1+\left( 100-1 \right)\times 1 \right]=50\left[ 2+99 \right]=50\times 101=5050$

(ii)

$\displaystyle {{S}_{n}}=1+2+3+...+n$

Here, $\displaystyle a=1,\,\,d=2-1=1,\,\,\,n=n$

$\displaystyle {{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$

= $\displaystyle \frac{n}{2}\left[ 2\times 1+\left( n-1 \right)\times 1 \right]$

= $\displaystyle \frac{n}{2}\left[ 2+n-1 \right]$$\displaystyle =\frac{n\left( n+1 \right)}{2}$

### Solution:

Using $\displaystyle {{S}_{n}}=\frac{n}{2}\left( a+{{a}_{n}} \right),$ with n = 17

$\displaystyle {{S}_{17}}=187,\,\,\,{{a}_{17}}=-13$

\$\displaystyle 187=\frac{17}{2}\left( a-13 \right)$

Þ$\displaystyle a-13=\frac{187\times 2}{17}=22$

Þ $\displaystyle a=22+13=35$…(i)

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

Þ $\displaystyle {{a}_{17}}=a+\left( 17-1 \right)d$

Þ$\displaystyle -13=35+16d$ [from (i)]

Þ $\displaystyle 16d=-48$

Þ
d = 3

### Solution:

(i)

Given $\displaystyle {{a}_{12}}=37,\,\,\,d=3,$ find a and S12.

\$\displaystyle a+11d=37$

or $\displaystyle a+11\times 3=37$or a = 4

\ $\displaystyle {{S}_{12}}=\frac{12}{2}\left[ a+{{a}_{12}} \right]=6\left[ 4+37 \right]=6\times 41=246$

(ii)

Given $\displaystyle a=8,\,\,{{a}_{n}}=62,\,\,{{S}_{n}}=210,$ find n and d.

$\displaystyle {{S}_{n}}=\frac{n}{2}\left[ a+{{a}_{n}} \right]$

$\displaystyle 210=\frac{n}{2}\left[ 8+62 \right]=\frac{n}{2}\times 70=35n$

Þ$\displaystyle n=\frac{210}{35}=6$

Also

Þ$\displaystyle 8+5d=62$or$\displaystyle 5d=62-8=62-8=54$

\$\displaystyle d=\frac{54}{5}=10.8$

(iii)

Given $\displaystyle l=28,\,\,{{S}_{n}}=144,\,\,$n = 9, find a.

\

$\displaystyle \frac{2}{9}\times 144=a+28$ or 32 = a + 28

Þa = 32 28 = 4

### Solution:

Here, $\displaystyle a=5,\,\,\,\,l={{a}_{n}}=45,\,\,\,{{S}_{n}}=400$

$\displaystyle \because$ $\displaystyle {{S}_{n}}=\frac{n}{2}\left[ a+l \right]$

\$\displaystyle 400=\frac{n}{2}\left[ 5+45 \right]=25n$

Þ $\displaystyle n=\frac{400}{25}=16$

Also $\displaystyle l$ = 45

\an = $\displaystyle a+\left( n-1 \right)d=45$

Þ

$\displaystyle 15d=45-5=40$

Þ $\displaystyle d=\frac{40}{15}=\frac{8}{3}$

Hence, number of terms(n) $\displaystyle =16,\,\,d=\frac{8}{3}$

### Solution:

$\displaystyle {{S}_{7}}=\frac{7}{2}\left[ 2a+\left( 7-1 \right) \right]d=49$[Given]

Þ$\displaystyle \frac{2}{7}\times 49=\left[ 2a+6d \right]$

$\displaystyle 14=2a+6d$

or$\displaystyle 7=a+3d$…(i)

Also $\displaystyle {{S}_{17}}=\frac{17}{2}\left[ 2a+\left( 17-1 \right)d \right]=289$

$\displaystyle \frac{2}{17}\times 289=2a+16d$

$\displaystyle 34=2a+16d$

or$\displaystyle a+8d=17$…(ii)

Subtracting (i) from (ii), we get

$\displaystyle -5d=-10$or d = 2

Putting d = 2 in (i), we get a = 7 3 ´ 2 = 1

\$\displaystyle {{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]=\frac{n}{2}\left[ 2\times 1+\left( n-1 \right)2 \right]$

$\displaystyle =n+n\left( n-1 \right)=n+{{n}^{2}}-n={{n}^{2}}$

### Solution:

Positive integers less than 400 are

1, 2, 3, 4, 5, …., 399

Here, $\displaystyle a=1,\,\,\,d=2-1=1,\,\,\,{{a}_{n}}=399$

\ $\displaystyle {{a}_{n}}=399$

$\displaystyle a+\left( n-1 \right)d=399$

$\displaystyle 1+\left( n-1 \right)\times 1=399$

Þ$\displaystyle n-1=399-1=398$

or$\displaystyle n=398+1=399$

\ $\displaystyle {{S}_{399}}=\frac{399}{2}\left[ 2\times 1+\left( 399-1 \right)\times 1 \right]$

$\displaystyle =\frac{399}{2}\left[ 2+398 \right]=\frac{399}{2}\times 400=79800$

### Solution:

Odd numbers between 0 and 50 are

1, 3, 5, 7, …, 49

Here, $\displaystyle a=1,\,\,\,d=3-1=2,\,\,\,{{a}_{n}}=49$

\ $\displaystyle {{a}_{n}}=49$

$\displaystyle a+\left( n-1 \right)d=49$

$\displaystyle 1+\left( n-1 \right)2=49$

$\displaystyle 2\left( n-1 \right)=49-1=48$

Þ $\displaystyle n-1=24$ or n = 25

\ $\displaystyle {{S}_{25}}=\frac{25}{2}\left[ a+{{a}_{25}} \right]$

$\displaystyle =\frac{25}{2}\left[ 1+49 \right]=25\times 25$

$\displaystyle =625$

### Solution:

Here, $\displaystyle a=$Rs 200, d = Rs. 50, n = 30

\$\displaystyle {{S}_{30}}=\frac{30}{2}\left[ 2a+\left( 30-1 \right)d \right]$

$\displaystyle =15\left[ 2\times 200+29\times 50 \right]$

$\displaystyle =15\left[ 400+1450 \right]=15\times 1850$

= Rs 27750

\Delay of 30 days will cost Rs. 27750 as penalty to the contractor.

### Solution:

Total cost of the shop = Rs. 120000

Cash $\displaystyle =\frac{1}{2}\times 120000$ = Rs. 60000

Interest for 1st installment = Rs $\displaystyle \frac{60000\times 12\times 1}{100}=$ Rs. 7200

Interest for 2nd installment = Rs $\displaystyle \frac{(60000-5000)\times 12\times 1}{100}=$ Rs. 6600

……………………………………………………..

……………………………………………………..

So, the interest for 12 installments will be

7200, 6600, 6000, 5400, 4800, …., 600

Writing in the reverse order we have

600, 1200, 1800, …., 7200

Here, a = Rs. 600, d = Rs. (1200 600) = Rs. 600, $\displaystyle {{a}_{n}}\,\,\,\text{or}\,\,\,{{a}_{12}}$ = Rs. 7200

\Sum of interest paid $\displaystyle =\frac{12}{2}\left[ a+{{a}_{12}} \right]=6\left[ 600+7200 \right]=6\times 7800=$ Rs. 46800

So, total cost of the shop = Rs. (60000 + 60000) + Rs. (46800)

= 166800

### Solution:

Here, $\displaystyle a=20,\,\,\,\,d=-1,\,\,\,{{S}_{n}}=200$

$\displaystyle {{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$

$\displaystyle 200=\frac{n}{2}\left[ 2\times 20+\left( n-1 \right)\,\left( -1 \right) \right]$

$\displaystyle 400=n\left( 40-n+1 \right)$

$\displaystyle 400=40n-{{n}^{2}}+n$

$\displaystyle {{n}^{2}}-41n+400=0$

$\displaystyle {{n}^{2}}-25n-16n+400=0$

$\displaystyle n\left( n-25 \right)-16\left( n-25 \right)=0$

$\displaystyle \left( n-25 \right)\left( n-16 \right)=0$

either $\displaystyle n=25$ or $\displaystyle n=16$

(i) Taking $\displaystyle n=25$

The number of logs in 25th row is

$\displaystyle {{a}_{25}}=20+\left( 25-1 \right)\left( -1 \right)$ $\displaystyle =20-24=-4$

Which is not possible as number of logs can not be negative

\ $\displaystyle n=25$ is rejected

(ii) Taking $\displaystyle n=16$

The number of logs in 16th row is

$\displaystyle {{a}_{16}}=20+15\times \left( -1 \right)$ $\displaystyle =20-15=5$

\ 5 logs are placed in 16th row

### Solution:

Here,

Þ$\displaystyle {{a}_{n+1}}-{{a}_{n}}=2-3n-3-2+3n=-3$

Þd. = 3

\

Hence $\displaystyle {{S}_{25}}=\frac{25}{2}\left[ 2\times (-1)+(25-1)\times (-3) \right]$

$\displaystyle =\frac{25}{2}\left[ -2+24\times (-3) \right]$ $\displaystyle =\frac{25}{2}\left[ -74 \right]=25\times (-37)=-925$

### Solution:

Let d be the common difference of the A.P. Given that a = 1 and

According to Problem:

$\displaystyle \left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}} \right)=\frac{1}{3}\left( {{a}_{5}}+{{a}_{6}}+{{a}_{7}}+{{a}_{8}} \right)$

$\displaystyle 3\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}} \right)=\left( {{a}_{5}}+{{a}_{6}}+{{a}_{7}}+{{a}_{8}} \right)$…(i)

Adding $\displaystyle \left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}} \right)$ to both sides of (i), we get

$\displaystyle \left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}} \right)+3\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}} \right)=\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}}+{{a}_{5}}+{{a}_{6}}+{{a}_{7}}+{{a}_{8}} \right)$

Þ$\displaystyle 4\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}} \right)=\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}}+...+{{a}_{8}} \right)$

or$\displaystyle 4{{S}_{4}}={{S}_{8}}$

Now, $\displaystyle {{S}_{4}}=\frac{4}{2}\left[ 2\times 1+\left( 4-1 \right)d \right]=4+6d$

and $\displaystyle {{S}_{8}}=\frac{8}{2}\left[ 2\times 1+\left( 8-1 \right)d \right]=8+28d$

According to the Problem:

$\displaystyle 4\left( 4+6d \right)=8+28d$

Þ16 + 24d = 8 + 28d

or

### Solution:

Here $\displaystyle a=18,\,\,\,d=-2$

$\displaystyle {{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$

Þ

Þ$\displaystyle \frac{n}{2}\left( 36-2n+2 \right)=0$

Þ$\displaystyle n(19-n)=0$

Þ$\displaystyle n=0$or$\displaystyle 19-n=0$Þn = 19

Hence, 19 terms of the sequence 8, 16, 14…. should be taken so that their sum is zero.

### Solution:

Volume of concrete needed to make first step

=$\displaystyle \ell \times b\times h=\frac{1}{4}\times \frac{1}{2}\times 50$m3

= $\displaystyle \frac{25}{4}$m3

Volume of concrete needed to make second step

$\displaystyle =2\times \frac{1}{4}\times \frac{1}{2}\times 50$ m3

= $\displaystyle \frac{50}{4}$m3

Volume of concrete needed to make third step

$\displaystyle =3\times \frac{1}{4}\times \frac{1}{2}\times 50$m3

$\displaystyle =\frac{75}{4}$m3

Here $\displaystyle a=\frac{25}{4},\,\,\,\,d=\frac{50}{4}-\frac{25}{4}=\frac{25}{4}$

\Total volume of concrete needed to make such 15 steps

= $\displaystyle \frac{15}{2}\left[ 2\times \frac{25}{4}+\left( 15-1 \right)\times \frac{25}{4} \right]$

$\displaystyle =\frac{15}{2}\left[ \frac{50}{4}+\frac{350}{4} \right]=\frac{15}{2}\times \frac{400}{4}=750$m3.

### Solution:

$\displaystyle {{r}_{1}}$ = 0.5 cm

$\displaystyle {{r}_{2}}$ = 1.0 cm

$\displaystyle {{r}_{3}}$ = 1.5 cm

Length of Ist spiral = pr1 (circumference of semicircle)

= $\displaystyle \frac{22}{7}\times 0.5=\frac{22}{7}\times \frac{1}{2}=\frac{11}{7}$

Length of IInd spiral = pr2

= $\displaystyle \frac{22}{7}\times 1=\frac{22}{7}$

Length of IIIrd spiral = pr3

= $\displaystyle \frac{22}{7}\times \frac{3}{2}=\frac{33}{7}$

\ $\displaystyle d=\frac{22}{7}-\frac{11}{7}=\frac{11}{7}$

$\displaystyle a=\frac{11}{7},\,\,\,\,n=13$

Length of spiral up of 13 consecutive semicircles

$\displaystyle =\pi {{r}_{1}}+\pi {{r}_{2}}+\pi {{r}_{3}}+....+\pi {{r}_{13}}$

$\displaystyle {{S}_{13}}=\frac{13}{2}\left[ 2\times \frac{11}{7}+\left( 13-1 \right)\frac{11}{7} \right]$ $\displaystyle =\frac{13}{2}\left( \frac{22}{7}+\frac{132}{7} \right)$

$\displaystyle =\frac{13}{2}\times \frac{154}{7}$ = 143 cm$\displaystyle$

### Solution:

Let a be the first term and d the common difference

According to question

$\displaystyle a+2d+a+6d=6$

$\displaystyle 2a+8d=6$

$\displaystyle a+4d=3$

$\displaystyle a=3-4d$…(i)

$\displaystyle \left( a+2d \right)\left( a+6d \right)=8$[Given]

$\displaystyle \left( 3-4d+2d \right)\left( 3-4d+6d \right)=8$[from (i)]

Þ$\displaystyle \left( 3-2d \right)\left( 3+2d \right)=8$

Þ$\displaystyle 9-4{{d}^{2}}=8$

Þ$\displaystyle 4{{d}^{2}}=1$

Þ$\displaystyle {{d}^{2}}=\frac{1}{4}$ or $\displaystyle d=\pm \frac{1}{2}$

Taking $\displaystyle d=\frac{1}{2}$

Þ$\displaystyle a=3-4\times \frac{1}{2}=1$

\$\displaystyle {{S}_{16}}=\frac{16}{2}\left[ 2\times 1+\left( 16-1 \right)\frac{1}{2} \right]=8\left[ 2+\frac{15}{2} \right]=76$

Taking $\displaystyle d=\frac{-1}{2}$Þ$\displaystyle a=3-4\left( \frac{-1}{2} \right)=5$

\$\displaystyle {{S}_{16}}=\frac{16}{2}\left[ 2\times 5+\left( 16-1 \right)\left( -\frac{1}{2} \right) \right]=8\left[ 10-\frac{15}{2} \right]=20$

## Selection of Terms in A.P.

Sometimes certain number of terms in A.P. are required. The following ways of selecting terms are convenient.

### Number of terms common difference

3$\displaystyle a-d,a,a+d$d

4$\displaystyle a-3d,a-d,,a+d,a+3d$2d

5$\displaystyle a-2d,a-d,a,a+d,a+2d$d

6$\displaystyle a-5d,a-3d,a-d,a+d,a+3d,a+5d$2d

### Solution:

$\displaystyle \because$Each prize is Rs. 20 less than the next most valuable one

Þ
d = Rs. 20

\ The four prizes are:

$\displaystyle a-3d,\,\,\,a-d,\,\,\,a+d,\,\,\,a+3d$

Þ$\displaystyle a-3\times 20,a-20,a+20,a+3\times 20$[$\displaystyle \because$ d = 20]

Þ$\displaystyle a-60,a-20,a+20,a+60$

According to question, we have

$\displaystyle a-60+a-20+a+20+a+60=280$

$\displaystyle 4a=280$

Þa = 70

Hence the four prizes are:

70 60 = Rs.10; 70 20 = Rs. 50; 70 + 20 = Rs. 90; 70 + 60 = Rs. 130.

### Solution:

Let the number be $\displaystyle \left( a-d \right),a,\left( a+d \right)$. Then

Sum $\displaystyle =-3\Rightarrow \left( a-d \right)+a+\left( a+d \right)=-3$

…(i)

Now, Product = 8

Þ$\displaystyle \left( a-d \right)\left( a \right)\left( a+d \right)=8$

Þ$\displaystyle a\left( {{a}^{2}}-{{d}^{2}} \right)=8$

Þ$\displaystyle \left( -1 \right)\left( 1-{{d}^{2}} \right)=8$ [from (i)]

Þ$\displaystyle {{d}^{2}}=9\Rightarrow d=\pm 3$

If d = 3 the numbers are $\displaystyle -4,-1,\ 2.$ If d = -3, the numbers are 2, -1, -4.

Thus, the numbers are $\displaystyle -4,-1,\ 2$ or $\displaystyle 2,-1,-4.$

### Solution:

Let the four parts be $\displaystyle \left( a-3d \right),\left( a-d \right),\left( a+d \right)$ and $\displaystyle \left( a+3d \right)$. Then, sum = 32

Þ$\displaystyle \left( a-3d \right)+\left( a-d \right)+\left( a+d \right)+\left( a+3d \right)=32\Rightarrow 4a=32\Rightarrow a=8$

According to question

$\displaystyle \frac{\left( a-3d \right)\left( a+3d \right)}{\left( a-d \right)\left( a+d \right)}=\frac{7}{15}$

Þ$\displaystyle \frac{{{a}^{2}}-9{{d}^{2}}}{{{a}^{2}}-{{d}^{2}}}=\frac{7}{15}$$\displaystyle \left[ \left( x+y \right)\left( x-y \right)={{x}^{2}}-{{y}^{2}} \right]$

Þ$\displaystyle \frac{64-9{{d}^{2}}}{64-{{d}^{2}}}=\frac{7}{15}$

Þ$\displaystyle 128{{d}^{2}}=512\Rightarrow {{d}^{2}}=4\Rightarrow d=\pm 2$

Thus, the four parts are $\displaystyle a-3d,a-d,a+d$ and $\displaystyle a+3d$ i.e., 2, 6, 10, 14.

### Solution:

According to Problem:

$\displaystyle {{S}_{1}}$ = sum of n term of an A.P. with a = 1 and d = 1

Þ$\displaystyle {{S}_{1}}=\frac{n}{2}\left[ \,2\times 1+\left( n-1 \right)\times 1\, \right]=\frac{n}{2}\left( n+1 \right)$

$\displaystyle {{S}_{2}}=$sum of n terms of an A.P. with a = 1 and d = 2

Þ$\displaystyle {{S}_{2}}=\frac{n}{2}\left[ 2\times 1+\left( n-1 \right)\times 2\, \right]={{n}^{2}}$

$\displaystyle {{S}_{3}}=$ sum of n terms of an A.P. with a = 1 and d = 3

Þ$\displaystyle {{S}_{3}}=\frac{n}{2}\left[ 2\times 1+\left( n-1 \right)\times 3\, \right]=\frac{n}{2}\left( 3n-1 \right)$

Now, $\displaystyle {{S}_{1}}+{{S}_{3}}=\frac{n}{2}\left( n+1 \right)+\frac{n}{2}\left( 3n-1 \right)=\frac{n}{2}(n+1+3n-1)=\frac{n}{2}\times 4n=2{{n}^{2}}$

and $\displaystyle {{S}_{2}}={{n}^{2}}$

Hence, $\displaystyle {{S}_{1}}+{{S}_{3}}=2{{S}_{2}}$

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