Answered

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The sum of the first 15 terms of an arithmetic series
IS 615
Given that
the fifth
term is 53 find the common difference

Sagot :

aayu0302

Answer:

d=4

Step-by-step 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

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