Question

pls answer this fast​

Answer 1

Answer:

1. S1

2. 111

hope it helpss you..

Answer 2

answer of question 2

The formular for the nth term of an arithmetic progression, Tn = a + (n - 1)d

where a is the first term and s the is the common difference.

T10 = a + (10 - 1)d

= a + 9d

since T10 = 31

a + 9d = 31 ----------- eqn(1)

T20 = a + (20 - 1)d

= a + 19d

since T20 = 71,

a + 19d = 71 ---------- eqn(2)

eqn(1) and eqn(2) form a simultaneous equation and can be solved by elimination method of solving simultaneous equation.

subtraction eqn(2) from eqn(1)

(a - a) + (9 - 19)d = (31 - 71)

-10d = -40

divide both sides by -10

d = 4

substitute d = 4 int eqn(1)

a + 9(4) = 31

a + 36 = 31

subtract 36 from both sides

a = 31 - 36

a = -5

Therefore, T30 = -5 + (30 - 1)4

= -5 + (29 x 4)

=-5 + 116

= 111

ans is option d) 111