Question

What is the sum of the first 12 terms of an arithmetic progression if the 3rd term is -13 and the 6th term is -4?

Answer 1

Answer:

3rd term :.............1                    6th term : ................2

a+2d= -13                                  a+5d= -4

subtracting 1 and 2 we get,

3d=9

d=3

therefore

a+2*3=-13

a=-19

hence sum of 12 terms =

S12= 12/2(2*-19+ (12-1)3)

S12= 6(-38+33)

therefore sum of 12 terms is equal to -30

hope it helps you mate

Answer 2

-30 is sum of 12th term

Step-by-step explanation:

Given that 3rd term = -13 and 6th term = -4

$a+2d=-13-(i)\\a+5d=-4-(ii)$

Subtracting equation (i) from (ii) we get,

$-3d=-9\\d=3$

Put $d=3$ in equation (i)

$a+2\times3 =-13\\a+6=-13\\a=-13-6\\a=-19$

to find sum of 12th term

$sn=\frac{n}{2}[2a+(n-1)d]\\sn=\frac{12}{2}[2\times(-19)+(12-1)3]\\=6[-38+(11)\times3]\\=6[-38+33]\\=6[-5]\\=-30$