1.

(a) In the formula `S_(n)=(n)/(2){2a+(n-1)d}`, make d as the subject. (b) Find the value of d, when `S_(n)=240,n=10,` and `a=6`.

Answer» (a) Given formula is `S_(n)=(n)/(2){2a+(n-1)d}`.
`rArr(n)/(2){2a+(n-1)d}=S_(n)`
`rArr 2a+(n-1)d=(2S_(n))/(n)`
`rArr (n-1)d=(2S_(n))/(n)-2a`
`rArr d=(2)/(n-1)[(S_(n))/(n)-a]`
(b) Substituting the values `S_(n)=240,n=10`, and a=6, we get.
`d=(2)/((10-1))[(240)/(10)-6]=(2)/(9)xx18=4`
`therefore d=4`.


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