An AP has 15 terms and a common difference of - 3. Find its first and last terms
if its sum is
b 15
d -120
a 120

Step-by-step explanation:

Sum =n/2{2a+(n-1)d

When sum=15 and d=-3

15=15/2{2a+(15-1)-3}

15=7.5{2a+(14)-3}

15=7.5{2a+(-42)}

15=7.5(2a-42)

15=15a-315

15+315=15a

330=15a

a=330/15

=22

a=22

Last term=a+(n-1)d

=22+(15-1)-3

=22+(14)-3

=22-42

= -20

When sum=120

Sum=n/2{2a+(n-1)d

120=15/2{2a+(15-1}-3

120=7.5{2a+(14)-3}

120=7.5{2a+(-42)}

120=7.5(2a-42)

120=15a-315

120+315=15a

435=15a

a=435/15

=29

a=29

Last term=a+(n-1)d

=29+(15-1)-3

=29+(14)-3

=29-42

= -13

When sum= -120

Sum=n/2{2a+(n-1)d

-120=15/2{2a+(15-1)-3

-120=7.5{2a+(14)-3}

-120=7.5{2a+(-42)}

-120=7.5(2a-42)

-120=15a-315

-120+315=15a

195=15a

a=195/15

a=13

Last term=a+(n-1)d

=13+(15-1)-3

=13-42

= -29