Question

Find the sum of 1+3+5+...+75

Answer 1

Answer:

Step-by-step explanation:a=1. d=2.

Tn =75

75=a+(n-1)d

75=1+(n-1)2

75=1+2n-2

75-1+2=2n

76/2=n

38=n

S38=n/2(2a+(n-1)d

=38/2(2x1+(38-1)2)

=19(2+37*2)

=19(2+74)

19*76

1444

Answer 2

Answer- The above question is from the chapter 'Arithmetic Progressions'.

Concept used: 1) Common difference (d) = a₂ - a₁

2) aₙ = a + (n - 1)d

where aₙ = nth term or last term of an AP

a = first term of AP

n = Number of terms of AP

d = Common difference

3) Sum of n terms (Sₙ) = $\frac{n}{2} [a + a_{n}]$

Given question: Find the sum of 1 + 3 + 5 +...+ 75.

Solution: Given series:- 1 + 3 + 5 +...+ 75.

a₁ = 1

a₂ = 3

d₁ = a₂ - a₁ = 3 - 1 = 2

a₃ = 5

d₂ = a₃ - a₂ = 5 - 3 = 2

Since d is same throughout, given series is an A.P. whose first term (a) = 1 and last term (aₙ) = 75 with common difference (d) = 2.

We know that, aₙ = a + (n - 1)d.

Substituting the values, we get,

75 = 1 + (n - 1)2

74/2 = n - 1

n (Number of terms) = 38

Sₙ = 1 + 3 + 5 +...+ 75

Sₙ = $\frac{n}{2} [a + a_{n}]$

Sₙ = $\frac{38}{2} [1 \: + \: 75]$

Sₙ = 1444

∴ sum of 1 + 3 + 5 +...+ 75 = 1444.