Question

Find the sum of 30 consecutive odd no. Who's last 127

Answer 1

$Assume\ the\ 30\ consecutive\ odd\ nos.\ are\ in\ AP. \\ \\ T_{30} = 127 \\ \\ d = 2\ (Odd\ nos.\ are\ added\ by\ 2) \\ \\ \\ T_{30} - T_1 = (30 - 1)d \\ \\ = 127 - T_1 = 29 \times 2 \\ \\ = 127 - T_1 = 58 \\ \\ \\ T_1 = 127 - 58 = 69$

$\\ \\ \\ 69\ is\ the\ first\ term. \\ \\ 69,\ 71,\ 73,.........,\ 123,\ 125,\ 127\ are\ the\ odd\ nos. \\ \\ \\ S_{30} \\ \\ = \frac{30}{2}[T_1 + T_{30}] \\ \\ = 15[69 + 127] \\ \\ = 15 \times 196 = \bold{2940}$

$\\ \\ \\ \therefore\ \bold{2940}\ is\ the\ answer.$

$\\ \\ \\ Hope\ this\ may\ be\ helpful. \\ \\ Please\ mark\ my\ answer\ as\ the\ \bold{brainliest}\ if\ this\ may\ be\ helpful. \\ \\ Thank\ you.\ Have\ a\ nice\ day. \\ \\ \\ \#adithyasajeevan$

Answer 2

Answer:

2940

Step-by-step explanation:

according to question,

n= 30

d= 2

an= 127

an= a+(n-1) d

127= a+ 29 * 2

after solving we get a = 69

s30 = 30/2 (69 +127)

s30 = 15 * 196

s30 = 2940

hope it helps you....