Let 'S' be the set of odd multiples of 7. i.e S={7,21,35,49,...}
where Tn=7(2n-1)
Range of a set is the difference between the largest and smallest numbers in the set. For set 'S' that contain 'n' elements, the smallest number is 7=T1
$$or\left[T1=7[2(1-1\right]=\ 7$$
$$l\arg est\ number\ is\ Tn\ =\ 7\left(2n-1\right)$$
$$Therefore,\ Tn-T1=7\left(2n-1\right)-7$$
$$=7\left(2n-2\right)$$
$$=7\left(2\right)\left(n-1\right)=14\left(n-1\right)$$
$$The\ value\ of\ \left(Tn-T1\right)\ is\ always\ a\ even\ multiple\ of\ 7.\ i.e$$
$$Tn-T1=\left[7\left(n-1\right)\right]2\ \ \ \ \left(multiples\ of\ 2\ are\ even\right)$$
$$from\ the\ options\ given,\ only\ '70'\ is\ an\ even\ multiple\ of\ '7'$$
$$Answer\ =\ 70....\ Option\ E$$