Question

Q1. Without adding, find the sum. (a) 1+3+5+....+31​

Answer 1

Answer: i) 1 + 3 + 5 + 7 + 9

Total consecutive odd numbers = 5

Thus, n = 5

Therefore, sum = n2

= 5 × 5 = 25

ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19

Total consecutive odd numbers = 10

Thus n = 10

Therefore sum = n × n

= 10× 10 = 100

iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

Total consecutive odd numbers = 12

Thus n = 12

Therefore sum = n × n

= 12× 12 = 144

Step-by-step explanation:

stay bless

Answer 2

Answer:

256

Step-by-step explanation:

This forms an Arithmetic Progression(AP) where,

first term, a = 1

Common difference, d = 3-1 = 2

and last term aₙ = 31

We need to find the number of terms first before finding the sum.

aₙ = 31

a + (n-1)d = 31     ---- we used the general formula aₙ = a + (n-1)d

1 + (n-1)2 = 31

(n-1)2 = 31 - 1

(n-1)2 = 30

(n-1) = 30/2

(n-1) = 15

n = 16

Sum to n terms is given by,

Sₙ = $\frac{n}{2} (a + aₙ)$

= $\frac{16}{2}$(1 + 31)

= 8 * 32 = 256