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_n=½n[2a+(n-1)d]
S_n=½n[2⋅1+(n-1)⋅1]
S_n=½n[2+(n-1)]
S_n=½n(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=½n[2a+(n-1)d] we get

S₂₃=½⋅23⋅[2⋅16+(23-1)(-5)]
=23⋅½⋅(32-110)
=23⋅(-78)=-897.
Also S_n=½n[2⋅16+(n-1)(-5)]=½n[32-5n+5]=½n[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_n=½n[2a+(n-1)d]
S_n=½n[2⋅2+(n-1)⋅2]
S_n=½n[4+2n-2]
S_n=½n(2n+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 nth term is (5-6n). Hence, find the sum of its first 20 terms.
Solution:
Let a_n be the nth term of the AP.
∴ a_n=5-6n
Putting n=1, we get
First term, a=a₁=5-6⋅1=-1
Putting n=2, we get

a₂=5-6⋅2=-7
Let (1 be the common difference of the AP.
∴ d=a₂–a₁=-7-(-1)=-7+1=-6.
Sum of first n term of the AP, S_n
S_n=½n[2a+(n-1)d]
S_n=½n[2∙(-1)+(n-1)∙(-6)]
=½∙n(-2-6n+6)
=n(2-3n)
=2n-3n²
Putting n=20, we get
S₂₀=2⋅20-3⋅20²=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_n=½n[2a+(n-1)d]
S₇=½⋅7⋅(2a+6d)=7⋅(a+3d)
S₁₇=½⋅17⋅(2a+16d)=17⋅(a+8d)
However, S₇=49 and S₁₇=289.
Now, 7(a+3d)=49
a+3d=7 … (i)
Also, 17(a+8d)=289
a+8d=17 … (ii)
Subtracting (i) from (ii), we ge:
=5d=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
=½n[2⋅1+(n-1)⋅2]=n[1+(n-1)]=n²
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, S₇=49 and S₁₇=289.
The sum of n terms of an AP is given by
S_n=½n[2a+(n-1)d]
S₇=½⋅7⋅[2a+6d]
49=7⋅(a+3d)
7=a+3d
a=7-3d … (1)
and S₁₇=½⋅17⋅[2a+16d]
289=17⋅(a+8d)
17=a+8d
Putting the value of a from equation (1), we get
17=7-3d+8d
5d=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_n=½n[2a+(n-1)d]
=½n[2⋅1+(n-1)⋅2]
=½n(2+2n-2)=n²