last term is term(n)
= a + (n-1)d
= 46
so -4 + nd - d = 46
d(n-1) = 50
sum(15) = (15/2)(2a + (n-1)d )
= (15/2)(-8 + 50) = 315
find the sum of first 15 terms of an A.P whose 1st and last are -4 and 46 respectively.
3 answers
Sn = n/2 (a1 + an) = 15/2(-4 + 46) = 15/2 * 42 = 315
I was going to use the formula
sum(n) = (n/2)(first + last), but then hesitated because we would
assume that there were 15 terms and term(15) = 46
So I went with the more general sum of terms formula
which contained d(n-1) as a constant of 50, so we got the same result.
however, suppose we had 26 terms, then term(26) = 46, and d = 2 , a = -4
sum(15) = (15/2)(-8 + 25(2)) = 315
suppose we have 21 terms then term(21) = 46 and d = 2.5
sum(15) = (15/2)(-8 + 20(2.5)) = 315
interesting
sum(n) = (n/2)(first + last), but then hesitated because we would
assume that there were 15 terms and term(15) = 46
So I went with the more general sum of terms formula
which contained d(n-1) as a constant of 50, so we got the same result.
however, suppose we had 26 terms, then term(26) = 46, and d = 2 , a = -4
sum(15) = (15/2)(-8 + 25(2)) = 315
suppose we have 21 terms then term(21) = 46 and d = 2.5
sum(15) = (15/2)(-8 + 20(2.5)) = 315
interesting