Problem 1. Find the sum of the first *n* natural numbers.

Solution:

The first *n* natural numbers are 1, 2, 3, 4, 5, …, *n*.

Here, *a*=1 and *d*=(2-1)=1.

Sum of *n* terms of an AP is given by

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

*S*=½

_{n}*n*[2⋅1+(

*n*-1)⋅1]

*S*=½

_{n}*n*[2+(

*n*-1)]

*S*=½

_{n}*n*(

*n*+1)

Example 7: Find the sum of 23 terms and *n* terms of the A.P. 16, 11, 6, 1 …

Solution: Here *a*=16, *a*+*d*=11 Therefore *d*=-5

Since *S _{n}*=½

*n*[2

*a*+(

*n*-1)

*d*] we get

*S*

_{23}=½⋅23⋅[2⋅16+(23-1)(-5)]

=23⋅½⋅(32-110)

=23⋅(-78)=-897.

Also

*S*=½

_{n}*n*[2⋅16+(

*n*-1)(-5)]=½

*n*[32-5

*n*+5]=½

*n*[37-5n]

P2. Find the sum of first *n* even natural numbers.

Solution:

The first *n* even natural numbers are 2, 4, 6, 8, 10, …, *n*.

Here, *a*=2 and *d*=(4-2)=2.

Sum of *n* terms of an AP is given by

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

*S*=½

_{n}*n*[2⋅2+(

*n*-1)⋅2]

*S*=½

_{n}*n*[4+2

*n*-2]

*S*=½

_{n}*n*(2

*n*+2)

*S*=

_{n}*n*(

*n*+1)

Hence, the required sum is

*n*(

*n*+1).

P3. Find the sum of first *n* terms of an AP whose *n*th term is (5-6*n*). Hence, find the sum of its first 20 terms.

Solution:

Let *a _{n}* be the

*n*th term of the AP.

∴

*a*=5-6

_{n}*n*

Putting

*n*=1, we get

First term,

*a*=

*a*

_{1}=5-6⋅1=-1

Putting

*n*=2, we get

*a*

_{2}=5-6⋅2=-7

Let (1 be the common difference of the AP.

∴

*d*=

*a*

_{2}–

*a*

_{1}=-7-(-1)=-7+1=-6.

Sum of first

*n*term of the AP,

*S*

_{n}*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

*S*=½

_{n}*n*[2∙(-1)+(

*n*-1)∙(-6)]

=½∙

*n*(-2-6

*n*+6)

=

*n*(2-3

*n*)

=2

*n*-3

*n*

^{2}

Putting

*n*=20, we get

*S*

_{20}=2⋅20-3⋅20

^{2}=40-1200=-1160

P4. The sum of the first 7 terms of an AP is 49 and the sum of its first 17 terms is 289. Find the sum of its first *n* terms.

Solution:

Let *a* be the first term and *d* be the common difference of the given AP. Then, we have:

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

*S*

_{7}=½⋅7⋅(2

*a*+6

*d*)=7⋅(

*a*+3

*d*)

*S*

_{17}=½⋅17⋅(2

*a*+16

*d*)=17⋅(

*a*+8

*d*)

However,

*S*

_{7}=49 and

*S*

_{17}=289.

Now, 7(

*a*+3

*d*)=49

*a*+3

*d*=7 … (i)

Also, 17(

*a*+8

*d*)=289

*a*+8

*d*=17 … (ii)

Subtracting (i) from (ii), we ge:

*d*=10

*d*=2

Putting

*d*=2 in (i), we get

*a*+6=7

*a*=1

Thus,

*a*=1 and

*d*=2.

∴ Sum of

*n*terms of AP

*n*[2⋅1+(

*n*-1)⋅2]=

*n*[1+(

*n*-1)]=

*n*

^{2}

Let’s read a particular post Derivation of the partial sum formula of every Arithmetic Series.

P5. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first

*n*terms.

Solution:

Here,

*S*

_{7}=49 and

*S*

_{17}=289.

The sum of

*n*terms of an AP is given by

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

*S*

_{7}=½⋅7⋅[2

*a*+6

*d*]

49=7⋅(

*a*+3

*d*)

7=

*a*+3

*d*

*a*=7-3

*d*… (1)

and

*S*

_{17}=½⋅17⋅[2

*a*+16

*d*]

289=17⋅(

*a*+8

*d*)

17=

*a*+8

*d*

Putting the value of

*a*from equation (1), we get

*d*+8

*d*

5

*d*=10

*d*=2

Putting the value of

*d*in equation (1), we get

*a*=7-3⋅2=1

The sum of

*n*terms of an AP is given by

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]

=½

*n*[2⋅1+(

*n*-1)⋅2]

=½

*n*(2+2

*n*-2)=

*n*

^{2}