1+2+3+3+4+...+97+98+99+100=
Answers
EXPLANATION.
Method = 1.
⇒ Series : 1 + 2 + 3 + . . . . . + 100.
As we know that,
Formula of :
Sum of n terms in A.P.
⇒ Sₙ = n/2[2a + (n - 1)d].
First term = a = 1.
Common difference = d = b - a = 2 - 1 = 1.
⇒ S₁₀₀ = 100/2[2(1) + (100 - 1)(1)].
⇒ S₁₀₀ = 50[2 + 99].
⇒ S₁₀₀ = 50 x 101.
⇒ S₁₀₀ = 5050.
Method = 2.
⇒ Series : 1 + 2 + 3 + . . . . . + 100.
As we know that,
Formula of :
Summation of n terms ∑n = [n(n + 1)/2].
⇒ ∑1 + 2 + 3 + . . . . . + 100.
⇒ [100(100 + 1)/2].
⇒ [100 x 101/2].
⇒ 5050.
MORE INFORMATION.
Supposition of terms in an A.P.
(1) = Three terms as : a - d, a, a + d.
(2) = Four terms as : a - 3d, a - d, a + d, a + 3d.
(3) = Five terms as : a - 2d, a - d, a, a + d, a + 2d.
Step-by-step explanation:
EXPLANATION.
Method = 1.
⇒ Series : 1 + 2 + 3 + . . . . . + 100.
As we know that,
Formula of :
Sum of n terms in A.P.
⇒ Sₙ = n/2[2a + (n - 1)d].
First term = a = 1.
Common difference = d = b - a = 2 - 1 = 1.
⇒ S₁₀₀ = 100/2[2(1) + (100 - 1)(1)].
⇒ S₁₀₀ = 50[2 + 99].
⇒ S₁₀₀ = 50 x 101.
⇒ S₁₀₀ = 5050.
Method = 2.
⇒ Series : 1 + 2 + 3 + . . . . . + 100.
As we know that,
Formula of :
Summation of n terms ∑n = [n(n + 1)/2].
⇒ ∑1 + 2 + 3 + . . . . . + 100.
⇒ [100(100 + 1)/2].
⇒ [100 x 101/2].
⇒ 5050.
MORE INFORMATION.
Supposition of terms in an A.P.
(1) = Three terms as : a - d, a, a + d.
(2) = Four terms as : a - 3d, a - d, a + d, a + 3d.
(3) = Five terms as : a - 2d, a - d, a, a + d, a + 2d.