Question

Find the sum of first 16 terms of an Arithmetic Progression whose
4th and 9th terms are - 15 and - 30 respectively.​

Answer 1

forth term = -15

a+3d= -15. -(1)

ninth term = -30

a+8d= -30. -(2)

on subtracting (1) from (2)-

we get , d= -3

on putting in (1)

we get , a= -6

Sn=n/2(2a+(n-1)d)

Sn=16/2(-12+(15)-3)

Sn=8(-12-45)

Sn=456