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45 votes
45 votes
The sum of the first 15 terms of an arithmetic series

IS 615
Given that
the fifth
term is 53 find the common difference

User Btav
by
2.7k points

1 Answer

21 votes
21 votes

Answer:

d=4

Explanation:

SUM OF 15 terms = 615

T5 = 53

T5 = a+(n-1)d

53 = a + 4d

4d = 53 - a

d = (53-a)/4

S15 = 15/2 (2a+14d) =615

15/2 [2a + 14×(53-a)/4 ]

=15/2 [ 2a +(53×14 - 14a )/4 ]

=15/2 [ (8a +742 - 14a ) /4]

615×8 = 15(-6a +742)

-6a+72=615×8/15 =41

-6a+72=42

-6a = - 72 + 42 = -30

a = 30/6= 5

a = 5

T5 = 5 + 4×d = 53

4d= 53 - 5 = 48

4d=48

d=4

Common difference is 4

User John Palmer
by
2.7k points