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