**Sum Of The First n Terms Of An Arithmetic Progression**

The sum of first n terms of an A.P. is given by

S_{n} = n/2 [2a + (n – 1) d] or S_{n} = n/2 [a + T_{n}]

**Note:**

**(i)** If sum of n terms Sn is given then general term T_{n} = S_{n} – S_{n-1} where S_{n-1} is sum of (n – 1) terms of A.P.

**(ii)** n^{th} term of an AP is linear in ‘n’

**Example:** a_{n} = 2 – n, a_{n} = 5n + 2 ……..

Also we can find common difference ‘d’ from an or T_{n} : d = coefficient of n

For a_{n} = 2 – n

∴ d = –1

**Verification:** by putting n = 1, 2, 3, 4,………

we get AP: 1, 0, –1, –2,……..

∴ d = 0 – 1 = –1

& for a_{n} = 5n + 2

d = 5

**(iii)** Sum of n terms of an AP is always quadratic in ‘n’

**Example: **S_{n} = 2n^{2} + 3n.

**Example: **S_{n} = n/4 (n + 1)

we can find ‘d’ also from S_{n}.

d = 2 (coefficient of n^{2})

for eg. : 2n^{2} + 3n, d = 2(2) = 4

**Verification:** S_{n} = 2n^{2} + 3n

at n = 1, S_{1} = 2 + 3 = 5 = first term

at n = 2, S_{2} = 2(2)^{2} + 3(2)

= 8 + 6 = 14 ≠ second term

= sum of first two terms.

∴ second term = S_{2} – S_{1} = 14 – 5 = 9

∴ d = a^{2} – a1 = 9 – 5 = 4

\( \text{Example: }{{S}_{n}}~=~\frac{n}{4}\left( n+1 \right) \)

\( {{S}_{n}}~=~\frac{{{n}^{2}}}{4}+\frac{n}{4} \)

\( \therefore d=2\left( \frac{1}{4} \right)=\frac{1}{2} \)

**Sum Of The n Terms Of An Arithmetic Progression With Examples**

**Example 1: **The sum of three numbers in A.P. is –3, and their product is 8. Find the numbers.

**Solution.** Let the numbers be (a – d), a, (a + d). Then,

Sum = – 3

⇒ (a – d) + a (a + d) = – 3

⇒ 3a = – 3

⇒ a = – 1

Product = 8

⇒ (a – d) (a) (a + d) = 8

⇒ a (a^{2} – d^{2}) = 8

⇒ (–1) (1 – d^{2}) = 8

⇒ d^{2} = 9 ⇒ d = ± 3

If d = 3, the numbers are –4, –1, 2. If d = – 3, the numbers are 2, – 1, –4.

Thus, the numbers are –4, –1, 2, or 2, – 1, – 4.

**Example 2: **Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.

**Solution.** Let the numbers be (a – 3d), (a – d), (a + d), (a + 3d), Then

Sum = 20

⇒ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20

⇒ 4a = 20

⇒ a = 5

Sum of the squares = 120

(a – 3d)^{2} + (a – d)^{2} + (a + d)^{2} + (a + 3d)^{2} = 120

⇒ 4a^{2} + d^{2} = 120

⇒ a^{2} + 5d^{2} = 30

⇒ 25 + 5d^{2} = 30 [∵ a = 5]

⇒ 5d^{2} = 5 ⇒ d = ± 1

If d = 1, then the numbers are 2, 4, 6, 8.

If d = – 1, then the numbers are 8, 6, 4, 2. Thus, the numbers are 2, 4, 6, 8 or 8, 6, 4, 2.

**Example 3: **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 (a – 3d), (a – d), (a + d) and (a + 3d). Then,

Sum = 32

⇒ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 32

⇒ 4a = 32 ⇒ a = 8

It is given that

\( \frac{(a-3d)\,(a+3d)}{(a-d)\,(a+d)}=\frac{7}{15} \)

\( \frac{{{a}^{2}}-9{{d}^{2}}}{{{a}^{2}}-{{d}^{2}}}=\frac{7}{15}\text{ }\Rightarrow \text{ }\frac{64-9{{d}^{2}}}{64-{{d}^{2}}}=\frac{7}{15} \)

⇒ 128d^{2} = 512

⇒ d^{2} = 4 ⇒ d = ± 2

Thus, the four parts are a – d, a – d, a + d and

a + 3d i.e. 2, 6, 10 and 14.

**Example 4: **Find the sum of 20 terms of the A.P. 1, 4, 7, 10, ……

**Solution. ** Let a be the first term and d be the common difference of the given A.P. Then, we have a = 1 and d = 3.

We have to find the sum of 20 terms of the given A.P.

Putting a = 1, d = 3, n = 20 in

S_{n} = \(\frac { n }{ 2 }\) [2a + (n – 1) d], we get

S_{20} = \(\frac { 20 }{ 2 }\) [2 × 1 + (20 – 1) × 3]

= 10 × 59 = 590

**Example 5: **Find the sum of first 30 terms of an A.P. whose second term is 2 and seventh term is 22.

**Solution.** Let a be the first term and d be the common difference of the given A.P. Then,

a_{2} = 2 and a_{7} = 22

⇒ a + d = 2 and a + 6d = 22

Solving these two equations, we get

a = – 2 and d = 4.

S_{n} = \(\frac { n }{ 2 }\) [2a + (n – 1) d]

∴ S_{30} = \(\frac { 30 }{ 2 }\) [2 × (–2) + (30 – 1) × 4]

⇒ 15 (–4 + 116) = 15 × 112

= 1680

Hence, the sum of first 30 terms is 1680.

**Example 6: **Find the sum of all natural numbers between 250 and 1000 which are exactly divisible by 3.

**Solution.** Clearly, the numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258, …., 999. This is an A.P. with first term

a = 252, common difference = 3 and last term = 999. Let there be n terms in this A.P. Then,

⇒ a_{n} = 999

⇒ a + (n – 1)d = 999

⇒ 252 + (n – 1) × 3 = 999 ⇒ n = 250

∴ Required sum = S_{n} = \(\frac { n }{ 2 }\) [a + l]

= \(\frac { 250 }{ 2 }\) [252 + 999] = 156375

**Example 7: **How many terms of the series 54, 51, 48, …. be taken so that their sum is 513 ? Explain the double answer.

**Solution. ** ∵ a = 54, d = – 3 and S_{n} = 513

⇒ \(\frac { n }{ 2 }\) [2a + (n – 1) d] = 513

⇒ \(\frac { n }{ 2 }\) [108 + (n – 1) × – 3] = 513

⇒ n^{2} – 37n + 342 = 0

⇒ (n – 18) (n – 19) = 0 ⇒ n = 18 or 19

Here, the common difference is negative, So, 19th term is a19 = 54 + (19 – 1) × – 3 = 0.

Thus, the sum of 18 terms as well as that of 19 terms is 513.

**Example 8: **If the m^{th} term of an A.P. is 1/n and the n^{th} term is 1/m, show that the sum of mn terms is (mn + 1).

**Solution. ** Let a be the first term and d be the common difference of the given A.P. Then,

\( {{a}_{m}}=\frac{1}{n}\Rightarrow a+(m-1)d=\frac{1}{n}\text{ }……\text{ (i)} \)

\( {{a}_{n}}=\frac{1}{n}\Rightarrow a+(n-1)d=\frac{1}{n}\text{ }……\text{ (ii)} \)

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

\( (m-n)d=\frac{1}{n}-\frac{1}{m} \)

\( \Rightarrow (m-n)d=\frac{m-n}{mn}\Rightarrow d=\frac{1}{mn} \)

Putting d = 1/mn in equation (i), we get

\( a+(m-1)\frac{1}{mn}=\frac{1}{n} \)

\( \Rightarrow a+\frac{1}{n}-\frac{1}{mn}=\frac{1}{n}\Rightarrow a=\frac{1}{mn} \)

\( Now,{{S}_{mn}}=\frac{mn}{2}\left\{ 2a+\left( mn1 \right)\times d \right\} \)

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

\( {{S}_{mn}}=\frac{1}{2}\left( mn+1 \right)~~ \)

**Example 9: **If the term of m terms of an A.P. is the same as the sum of its n terms, show that the sum of its (m + n) terms is zero.

**Solution. ** Let a be the first term and d be the common difference of the given A.P. Then,

S_{m} = S_{n}

⇒ \(\frac { m }{ 2 }\) [2a + (m – 1) d] = \(\frac { n }{ 2 }\) [2a + (n – 1) d]

⇒ 2a(m – n) + {m (m – 1) – n (n – 1)} d = 0

⇒ 2a (m – n) + {(m2 – n2) – (m – n)} d = 0

⇒ (m – n) [2a + (m + n – 1) d] = 0

⇒ 2a + (m + n – 1) d = 0

⇒ 2a + (m + n – 1) d = 0 [∵ m – n ≠ 0] ….(i)

\( {{S}_{m+n}}=\frac{m+n}{2}\left\{ 2a+\left( m+n-1 \right)d \right\} \)

\( {{S}_{m+n}}=\frac{m+n}{2}\times 0=0\text{ }\left[ \text{Using equation }\left( \text{i} \right) \right] \)

**Example 10: **The sum of n, 2n, 3n terms of an A.P. are S_{1}, S_{2}, S_{3} respectively. Prove that S_{3} = 3(S_{2} – S_{1}).

**Solution. ** Let a be the first term and d be the common difference of the given A.P. Then,

S_{1} = Sum of n terms

⇒ S_{1} = \(\frac { n }{ 2 }\) {2a + (n – 1)d} ….(i)

S_{2} = Sum of 2n terms

⇒ S_{2} = \(\frac { 2n }{ 2 }\) [2a + (2n – 1) d] ….(ii)

and, S_{3} = Sum of 3n terms

⇒ S_{3} = \(\frac { 3n }{ 2 }\) [2a + (3n – 1) d] ….(iii)

Now, S_{2} – S_{1}

= \(\frac { 2n }{ 2 }\) [2a + (2n – 1) d] – \(\frac { n }{ 2 }\) [2a + (n –1) d]

S_{2} – S_{1} = \(\frac { n }{ 2 }\) [2 {2a + (2n – 1)d} – {2a + (n – 1)d}]

= \(\frac { n }{ 2 }\) [2a + (3n – 1) d]

∴ 3(S_{2} – S_{1}) = \(\frac { 3n }{ 2 }\) [2a + (3n – 1) d] = S_{3 }[Using (iii)]

Hence, S_{3} = 3 (S_{2} – S_{1})

**Example 11: **The sum of n terms of three arithmetical progression are S_{1}, S_{2} and S_{3}. The first term of each is unity and the common differences are

1, 2 and 3 respectively. Prove that S_{1} + S_{3} = 2S_{2}.

**Solution.** We have,

S_{1} = Sum of n terms of an A.P. with first term 1 and common difference 1

= \(\frac { n }{ 2 }\) [2 × 1 + (n – 1) 1] = \(\frac { n }{ 2 }\) [n + 1]

S_{2} = Sum of n terms of an A.P. with first term 1 and common difference 2

= \(\frac { n }{ 2 }\) [2 × 1 + (n – 1) × 2] = n^{2}

S_{3} = Sum of n terms of an A.P. with first term 1 and common difference 3

= \(\frac { n }{ 2 }\) [2 × 1 + (n – 1) × 3] = \(\frac { n }{ 2 }\) (3n – 1)

Now, S_{1} + S_{3} = \(\frac { n }{ 2 }\) (n + 1) + \(\frac { n }{ 2 }\) (3n – 1)

= 2n^{2} and S_{2} = n^{2}

Hence S_{1} + S_{3} = 2S_{2}

**Example 12: **The sum of the first p, q, r terms of an A.P. are a, b, c respectively. Show that

\(\frac { a }{ p }\) (q – r) + \(\frac { b }{ q }\) (r – p) + \(\frac { c }{ r }\) (p – q) = 0

**Solution. ** Let A be the first term and D be the common difference of the given A.P. Then,

a = Sum of p terms ⇒ a = \(\frac { p }{ 2 }\) [2A + (q – 1) D]

⇒ \(\frac { 2a }{ p }\) = [2A + (p – 1) D] ….(i)

b = Sum of q terms

⇒ b = \(\frac { q }{ 2 }\) [2A + (q – 1) D]

⇒ \(\frac { 2b }{ q }\) = [2A + (q – 1) D] ….(ii)

and, c = Sum of r terms

⇒ c = \(\frac { r }{ 2 }\) [2A + (r – 1) D]

⇒ \(\frac { 2c }{ r }\) = [2A + (r – 1) D] ….(iii)

Multiplying equations (i), (ii) and (iii) by (q – r), (r – p) and (p – q) respectively and adding, we get

\(\frac { 2a }{ p }\) (q – r) + \(\frac { 2b }{ q }\) (r – p) + \(\frac { 2c }{ r }\) (p – q)

= [2A + (p – 1) D] (q – r) + [2A + (q – 1) D] (r – p) + [(2A + (r – 1) D] (p – q)

= 2A (q – r + r – p + p – q) + D [(p – 1) (q – r) + (q – 1)(r – p) + (r – 1) (p – q)]

= 2A × 0 + D × 0 = 0

**Example 13: **The ratio of the sum use of n terms of two A.P.’s is (7n + 1) : (4n + 27). Find the ratio of their m^{th} terms.

**Solution.** Let a_{1}, a_{2} be the first terms and d_{1}, d_{2} the common differences of the two given A.P.’s .Then the sums of their n terms are given by

S_{n} = \(\frac { n }{ 2 }\) [2_{1} + (n – 1) d_{1}], and S_{n}‘ = \(\frac { n }{ 2 }\) [2a_{2} + (n – 1) d_{2}]

\( \therefore \frac{{{S}_{n}}}{S_{n}^{‘}}=\frac{\frac{n}{2}[2{{a}_{1}}+(n-1){{d}_{1}}]}{\frac{n}{2}[2{{a}_{2}}+(n-1){{d}_{2}}]}=\frac{2{{a}_{1}}+(n-1){{d}_{1}}}{2{{a}_{2}}+(n-1){{d}_{2}}} \)

It is given that

\( \frac{{{S}_{n}}}{S_{n}^{‘}}=\frac{7n+1}{4n+27} \)

\( \Rightarrow \frac{2{{a}_{1}}+(n-1){{d}_{1}}}{2{{a}_{2}}+(n-1){{d}_{2}}}=\frac{7n+1}{4n+27} \)

To find the ratio of the mth terms of the two given A.P.’s, we replace n by (2m – 1) in equation (i). Then we get

\( \therefore \frac{2{{a}_{1}}+(2m-2){{d}_{1}}}{2{{a}_{2}}+(2m-2){{d}_{2}}}=\frac{7(2m-1)+1}{4(2m-1)+27} \)

\( \Rightarrow \frac{{{a}_{1}}+(m-1){{d}_{1}}}{{{a}_{2}}+(m-1){{d}_{2}}}=\frac{14m-6}{8m+23} \)

Hence the ratio of the m^{th} terms of the two A.P.’s is (14m – 6) : (8m + 23)

More examples on Arithmetic Progression

**Example 14: **The ratio of the sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of the m^{th} and n^{th} terms is (2m – 1) : (2n – 1).

**Solution.** Let a be the first term and d the common difference of the given A.P. Then, the sums of m and n terms are given by

S_{m} = \(\frac { m }{ 2 }\) [2a + (m – 1) d], and S_{n} = \(\frac { n }{ 2 }\) [2a + (n – 1) d]

respectively. Then,

\( \frac{{{S}_{m}}}{{{S}_{n}}}=\frac{{{m}^{2}}}{{{n}^{2}}}\Rightarrow \frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]}=\frac{{{m}^{2}}}{{{n}^{2}}} \)

\( \Rightarrow \frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m}{n} \)

⇒ [2a + (m – 1) d] n = {2a + (n – 1) d} m

⇒ 2a (n – m) = d {(n – 1) m – (m – 1) n}

⇒ 2a (n – m) = d (n – m)

⇒ d = 2a

\( \text{Now, }\frac{{{T}_{m}}}{{{T}_{n}}}=\frac{a+(m-1)d}{a+(n-1)d} \)

\( =\frac{a+(m-1)2a}{a+(n-1)2a}=\frac{2m-1}{2n-1} \)