# Calculate the Partial Sum of each Arithmetic Sequence or Series Given Two Terms

Problem 1. Find the sum of first 51 terms of an arithmetic sequence whose second and third terms are 14 and 18 respectively.
Solution:
Let a be the first term and d be the common difference of the arithmetic sequence. Then,

d=a3a2=18-14=4
Given a2=14.
[an=a+(n-1)d]
a+d=14
a+4=14
a=14-4=10

Using the formula, Snn[2a+(n-1)d], we get
S51=½⋅51⋅[2⋅10+(51-1)⋅4]
=½⋅51⋅(20+200)
=½⋅51⋅220
=5610.

Hence, the required sum is 5610.

P2. The 13th terms of an arithmetic sequence is 4 times its 3rd term. If its 5th term is 16, Find the sum of its first 10 terms.
Solution:
Let a be the first term and d be the common difference of the arithmetic sequence. Then, a13=4⋅a3 (Given).

[an=a+(n-1)d]
a+12d=4(a+2d)
a+12d=4a+8d
3a=4d … (1)
Also, a5=16 (Given)
a+4d=16 … (2)

Solving (1) and (2), we get
a+3a=16
4a=16
a=4

Putting a=4 in (1), we get
4d=3⋅4=12
d=3
Using the formula, Snn[2a+(n-1)d], we get
S10=½⋅10⋅[2⋅4+(10-1)⋅3]
=5⋅(8+27)
=5⋅35
=175.

Hence, the required sum is 175.

P3. The 16th term of an arithmetic sequence is 5 times its 3rd term. If its 10th term is 41, find the sum of its first 15 terms.
Solution:
Let a be the first term and d be the common difference of the arithmetic sequence. Then,

a6=5⋅a3 (Given)
[an=a+(n-1)d]
a+15d=5(a+2d)
a+15d=5a+10d
4a=5d … (1)

Also,
a10=41
a+9d=41
a=41-9d … (2)

Solving (1) and (2), we get
(1) … 4(41-9d)=5d
164-36d=5d
164=5d+36d
164=41d
d=4
(2) … a=41-9⋅4
=41-36
a=5

Using the formula, Snn[2a+(n-1)d], we get
S15=½⋅15⋅[2⋅5+(15-1)⋅4]
=15⋅½⋅(10+56)
=15⋅33
=495.

Hence, the required sum is 495.
Let’s read an important post Derivation of the partial sum formula of every Arithmetic Series.
P4. The 13th term of an arithmetic sequence is 4 times its 3rd term. If its 5th term is 16 then the sum of its first ten terms is
(a) 150 (b) 175 (c) 160 (d) 135
Solution: Let a be the first term and d be the common difference of the arithmetic sequence. Then,
Given a13=4a3.
[an=a+(n-1)d]
a+12d=4(a+2d)
a+12d=4a+8d
3a=4d … (1).
Given a5=16
a+4d=16 … (2)

Solving (1) and (2), we ge
a+3a=16
4a=16
a=4.

Putting a=4 in (1), we get
4d=3⋅4=12
d=3.

Using the formula, Snn[2a+(n-1)d]. we get
S10=½⋅10⋅[2⋅4+(10-1)⋅3]
=5⋅(8+27)
=5⋅35
=175.

Thus, the sum of its first 10 terms is 175.

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