SOLUTION: THE FIFTH TERM OF AN ARITHEMETIC SEQUENCE IS 10 & THE SUM OF THE THE FIRST 10 TERMS IS 115. FIND IT'S FIRST TERM AND IT'S COMMON DIFFERENCE.

Question 776837: THE FIFTH TERM OF AN ARITHEMETIC SEQUENCE IS 10 & THE SUM OF THE THE FIRST 10 TERMS IS 115. FIND IT'S FIRST TERM AND IT'S COMMON DIFFERENCE.
Answer by pakhi(24) (Show Source): You can put this solution on YOUR website!
Let the 1st term be 'a' and the common difference be 'b'.
Therefore the 5th term is a+4b = 10(given)----------------(1)
The formula for the sum of 15 terms of an A.P. is
S(10)= (10/2)(2a+9b)=115
or S(10)= 5(2a+9b)=115
or S(10)= 10a+45b =115 ----------------------------------(2)

Multiplying equation1 by 10 we have 10a+40b = 100 --------(3)
Subtracting equation3 from equation2 we have 5b= 15
So we have b=3(common difference)
Putting the value of 'b' in equation1 we have a+4*3=10
So a=10-12= -2(first term of the series)

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