Answer:
In an arithmetic sequence, whose first term is a and difference between a term and its preceding term is d,
the nth term is a+(nâ’1)d and sum of first n terms is n2(2a+(nâ’1)d)
Hence 6th term will be a+5d=8 and 10th term will be a+9d=13
Subtracting first from second, 4d=5 or d=1.25
and a=8â’5â‹…1.25=8â’6.25=1.75
Hence sum of first four terms is
42â‹…(2â‹…1.75+3â‹…1.25)=2â‹…(3.5+3.75)=2â‹…7.25=14.50
