Question

What is the sum of the first 36 terms of the arithmetic sequence in which `a_36` =14 and the common difference is `d= 1/8`?

Answer 1

Taking 36 th term `a_36=14` as first term and treating the arithmetic sequence in reverse way we will have common difference `d=-1/8`. Thus the sum of first 36 terms will be

`S=36/2[2xxa_36+(36-1)xxd]`

`=>S=36/2[2xx14+(36-1)xx(-1/8)]`

`=>S=36xx14+18xx35xx(-1/8)]`

`=>S=504-315/4=425 1/4`

Answer 2

`S_(36)=1701/4`

Explanation:

`"the nth term of an arithmetic sequence is"`

`•color(white)(x)a_n=a+(n-1)d`

`"where a is the first term and d the common difference"`

`"we can find a using "a_(36)=14`

`rArra+35d=14`

`rArra=14-(35xx1/8)=77/8`

`•color(white)(x)S_n=n/2[2a+(n-1)d]larrcolor(blue)"sum to n terms"`

`rArrS_(36)=18(77/4+35/8)=1701/4`