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*=

*a*

_{3}–

*a*

_{2}=18-14=4

Given

*a*

_{2}=14.

[

*a*=

_{n}*a*+(

*n*-1)

*d*]

*a*+

*d*=14

*a*+4=14

*a*=14-4=10

Using the formula,

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*], we get

*S*

_{51}=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, *a*_{13}=4⋅*a*_{3} (Given).

*a*=

_{n}*a*+(

*n*-1)

*d*]

*a*+12

*d*=4(

*a*+2

*d*)

*a*+12

*d*=4

*a*+8d

3

*a*=4

*d*… (1)

Also,

*a*

_{5}=16 (Given)

*a*+4

*d*=16 … (2)

Solving (1) and (2), we get

*a*+3

*a*=16

4

*a*=16

*a*=4

Putting

*a*=4 in (1), we get

*d*=3⋅4=12

*d*=3

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*], we get

*S*

_{10}=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,

*a*

_{6}=5⋅

*a*

_{3}(Given)

[

*a*=

_{n}*a*+(

*n*-1)

*d*]

*a*+15

*d*=5(

*a*+2

*d*)

*a*+15

*d*=5

*a*+10

*d*

4

*a*=5

*d*… (1)

Also,

*a*

_{10}=41

*a*+9

*d*=41

*a*=41-9

*d*… (2)

Solving (1) and (2), we get

*d*)=5d

164-36

*d*=5

*d*

164=5

*d*+36

*d*

164=41

*d*

*d*=4

(2) …

*a*=41-9⋅4

=41-36

*a*=5

Using the formula,

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*], we get

*S*

_{15}=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

*a*

_{13}=4

*a*

_{3}.

*a*=

_{n}*a*+(

*n*-1)

*d*]

*a*+12

*d*=4(

*a*+2

*d*)

*a*+12

*d*=4

*a*+8

*d*

3

*a*=4

*d*… (1).

Given

*a*

_{5}=16

*a*+4

*d*=16 … (2)

Answer: (b) 175

Solving (1) and (2), we ge

*a*+3

*a*=16

4

*a*=16

*a*=4.

Putting

*a*=4 in (1), we get

*d*=3⋅4=12

*d*=3.

Using the formula,

*S*=½

_{n}*n*[2

*a*+(

*n*-1)

*d*]. we get

*S*

_{10}=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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