SOLUTION: The 6th term of an AP is 35 and the 13term is 77calculate S6-S4?

Question 1178050: The 6th term of an AP is 35 and the 13term is 77calculate S6-S4?

Found 3 solutions by MathLover1, chhavi_1997, ikleyn:
Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!

if 6th term and 13term is , we have

.......solve for
.............eq.1

.......solve for
.............eq.2

equate right sides of eq.1 and eq.2

.....solve for

go to
.............eq.1, substitute

so, your n-th term formula is:

now calculate

since given 6th term

then

Answer by chhavi_1997(1)   (Show Source): You can put this solution on YOUR website!
Given t6=35
t13=77
therefore, t6= a+(n-1)d=a+(6-1)d
35=a+5d .....1
t13= a+(13-1)d= a+12d
77=a+12d ...2
by solving 1 and 2 we have
a=5, d=6
now, sn=n/2(2a+(n-1)d) using this formula
s6=6/2(2*5+(6-1)*6)
s6=3(10+5*6)
s6=120
similary,s4=4/2(2*5+(4-1)*6)
s4=2(10+3*6)
s4=56
therefore, s6-s4= 120-56
s6-s4=64

Answer by ikleyn(52787)   (Show Source): You can put this solution on YOUR website!
.

            The problem does not define the meaning of S6, S4, so I will suppose that
            S6 and S4 are the sums of the first 6 and the first 4 terms of the AP, respectively.

There are 13 - 6 = 7 gaps between the 13-th and 6-th terms of the AP on the number line, so each gap is = 6 units. Thus the common difference of the AP is d= 6. Now, - = + = + = (35-6) + 35 = 70-6 = 64. It is the answer to the problem's question: - = 64.

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    - The proofs of the formulas for arithmetic progressions
    - Problems on arithmetic progressions
    - Word problems on arithmetic progressions
    - Finding number of terms of an arithmetic progression
    - Inserting arithmetic means between given numbers
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