An arithmetic sequence has first term a and common difference d. The sum of the first 10 terms of the sequence is 162.

Question

a) Show that 10a+45d=162

Given that the sixth term of the sequence is 17.

b) Find the value of a and d.

Answer 1

the formula is S_n=(n/2)[2a+(n-1)d]

162=(10/2)[2a+(10-1)d]

162=(5)[2a+9d]

162=5*2a+5*9d

162=10a+45d

 

the formula is x_n=a+d(n-1)

17=a+d(6-1)

17=a+5d

17-5d=a

back to the answer for part a)

162=10a+45d

162=10(17-5d)+45d

162=170-50d+45d

162=170-5d

162-170=-5d

-8=-5d

d=8/5

17-5(8/5)=a

17-8=a

a=9

check: 162=(10/2)[2*9+(10-1)(8/5)]

162=5[18+9(8/5)]

162=5[18+14 2/5]

162=5[32 2/5]

162=160+2 (using the distributive property)

162=162