Question

.

Answer 1

u See it is a Ap with d=-3 t2-t1 -5 -(-2) = -3

Answer 2

\[Sn = \frac{ n }{ 2 }[2a+(n-1)d]\]

Answer 3

\[\rm S = {n \over 2}\left( a_1 + a_n\right)\]

Answer 4

n= no of term

a= 1st term

d=common Difference

Answer 5

and a^n equals ?

Answer 6

\[\rm S = {20 \over 2}\left(-2 + a_{20} \right)\]

Answer 7

Use \(\rm a_n = a_1 + (n - 1)d\)

Answer 8

ok

Answer 9

so an = 20?

Answer 10

the anwser is -100

Answer 11

\[\rm a_n = -2 + 19(-3) = -59\]

Answer 12

so the answer is -59

Answer 13

\[\rm S = {n \over 2}{(a_1 + a_20)} = {20 \over 2}(-2 - 59) = 10(-61)\]

Answer 14

-610

Answer 15

Yes.