Question

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

Answer 1

bro A4=a+3d=-15
A9=a+8d =-30
subtract both equations,
5d=-15,
D=-3.
then a+3(-3)=-15
a=-6.
A10=a+9d
=-6+9(-3)
=-33

then sum =n/2(first term+last term)
=10/2(-6+(-33))
=5×(-39)
=-195
I hope it helps you buddy

Answer 2

$the \: forth \: term \: is - 15 \\ ie \\ a + (4 - 1)d = - 15 \\ a + (9 - 1)d = - 30... \\ so \: on \: solving \: both equations \\ a + 3d = - 15 \\ a + 8d = - 30 \\ 3d - 8d = - 15 + 30 \\ - 5d = 15 \\ d = - 3 \\ and \: n by \: the1st \: equation \\ a + 3 \times - 3 = - 15 \\ a - 9 = - 15 \\ a = 5 \div 3 \\ so \: sum \: of \: 10 \: . \\ n \div 2 \times {2a + (n - 1)d} \\ = 10 \div 2 \times 2 \times 5 \div 3 + 9 \times - 3 \\ = 5 \times 10 \div 3 + 9 \times - 3 \\ = 50 \div 3 + - 27 \\ = (50 - 81) \div 3 \\ = - 31 \div 3$