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

 

Problem:

 

Find the sum of the first

(i) 100 natural numbers

(ii)n natural numbers.

 

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}

 

Problem:

 

The sum of the first 17 terms of an arithmetic sequence is 187. If \displaystyle {{a}_{17}}=-13, find a and d

 

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

 

Problem:

 

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

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

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

 

Solution:

 

(i)

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

\\displaystyle a+11d=37

or \displaystyle a+11\times 3=37or 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=62or\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

 

Problem:

 

The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

 

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}

 

Problem:

 

If the sum of 7 term of an A.P. is 49 and that of 17 term is 289, find the sum of n terms.

 

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=-10or 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}}

 

Problem:

 

Find the sum of positive integers less than 400.

 

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

 

Problem:

 

Find the sum of the odd numbers between 0 and 50.

 

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

 

Problem:

 

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc. the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much does a delay of 30 days cost the contractor?

 

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.

 

Problem:

 

Raghav buys a shop for Rs 120000. He pays half of the amount in cash and agrees to pay the balance in 12 annual installments of Rs. 5000 each. If the rate of interest is 12% and he pays with the instalment the interest due on the unpaid amount, find the total cost of the shop.

 

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

 

Problem:

 

200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (See figure). In how many rows the 200 logs are placed and how many logs are in the top row.


 

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

 

Problem:

 

Find the sum of the first 25 terms of an A.P. whose nth term is given by \displaystyle {{a}_{n}}=2-3n.

 

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

 

Problem:

 

Find the common difference of an A.P. whose first term is 1 and the sum of the first four term is one-third the sum of the next four terms.

 

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

 

Problem:

 

How many terms of the sequence 18, 16, 14, …. should be taken so that their sum is zero?

 

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=0or\displaystyle 19-n=0Þn = 19

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

 

Problem:

 

A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete.

Each step has a rise of \displaystyle \frac{1}{4}m and a tread of \displaystyle \frac{1}{2}m see figure Calculate the total volume of concrete required to built the terrace.


 

Solution:

 

Volume of concrete needed to make first step

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

= \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 50m3

\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}=750m3.

 

 

Problem:

 

A spiral is made up of successive semicircles, with centres alternatively at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? Take \displaystyle \left( \pi =\frac{22}{7} \right)


 

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

 

Problem:

 

The sum of the third and the seventh terms of an A.P. is 6 and their product is 8. Find the sum of first sixteen terms of the A.P.

 

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+dd

4\displaystyle a-3d,a-d,,a+d,a+3d2d

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

6\displaystyle a-5d,a-3d,a-d,a+d,a+3d,a+5d2d

 

Problem:

 

A sum of Rs 280 is to be used to award four prizes. If each prize after the first is Rs. 20 less than the next most valuable one, find the value of each of the prize.

 

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.

 

Problem:

 

The sum of three numbers in A.P. is -3, and their product is 8. Find the numbers.

 

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.

 

 

Problem:

 

Divide 32 into four parts which are in A.P. such that the product of extremes is to the product of means is 7:15.

 

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.

 

Problem:

 

The sums of n terms of three arithmetical progressions are \displaystyle {{S}_{1}},{{S}_{2}} and \displaystyle {{S}_{3}}. The first term of each is unity and the common differences are 1, 2, and 3 respectively. Prove that \displaystyle {{S}_{1}}+{{S}_{3}}=2{{S}_{2}}.

 

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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