Question

the ap in which 4 term is minus 15 and 9th term is minus 30 find the sum of first 10 numbers

Answer 1

given that,
a+3d= -15
=> a=-15-3d
and
a+8d=-30
=> -15-3d=-30-8d
=> 15=-5d
=> d=-3
therefore,
a= -6
Now,
S10= 10/2(-6-33)
S10= 5×(-39)=-195

Answer 2

Here is your solution

Given :-

The AP in which 4th term is -15
I. e
a4= (a+3d)= -15 ..................... (1)

9th term is -30
I.e
a9= (a+8d)= -30..................... (2)

on subtracting equation (1) from (2)

(a+3d) - (a +8d) = -15 + 30
-5d= 15
d=-15/5
d= -3

put in value of d in equation (1)

a+3d = -15
a-9= -15
a= -6

S10=?

Sn=n/2(2a+(n-1)d)
S10=10/2(2(-6)+(10-1)-3)
S10=5(-12-27)
S10=5(-39)
S10= -195

So, the sum of first 10 terms of this A.P. is -195,

Hope it helps