confusing wording.
I interpret that to say, d = -3
sum(15) = -120
sum(15) = (15/2)(2a + 14(-3) ) = -120
15( 2a - 42) = -240
2a - 42 = -16
2a = 26
a = 13
first term is 13
last term = term(15) = a+14d
= 13 + 14(-3) = -29
check:
sum(15) = (15/2)(first + last)
= (15/2)(13 -29) = -120
An ap has 15th and a common difference of -3 find the first and last term. If it. Sum is -120
7 answers
wow - pretty garbled language. I will assume you meant
An a.p. has 15 terms and a common difference of -3. Find the first and last term if its sum is -120 .
S15 = 15/2 (2a+14d)
15/2 (2a+14(-3)) = 120
a = 29
a+14d = -13
An a.p. has 15 terms and a common difference of -3. Find the first and last term if its sum is -120 .
S15 = 15/2 (2a+14d)
15/2 (2a+14(-3)) = 120
a = 29
a+14d = -13
Oops -- I missed the -120.
Good
How did u get ur 120
if it sum is 15 nko
Good