Question

See the value of 1 + 3 + 5 + …… .. + 999.​

Answer 1

Step-by-step explanation:

Given series is Arithmetic Progression as difference between consecutive terms is constant.

nth term of A.P. is,

a

n

=a

1

+(n−1)d

In given problem,

a

n

=999

a

1

=1

d=3−1=2

∴999=1+(n−1)×2

∴998=2(n−1)

∴n−1=

2

998

∴n−1=499

∴n=500

2) Sum of n terms of A.P. is

∑a

n

=

2

n

[2a+(n−1)d]

∴∑a

n

=

2

500

[(2×1)+(500−1)×2]

∴∑a

n

=250[2+(499×2)]

∴∑a

n

=250[2+998]

∴∑a

n

=250[1000]

∴∑a

n

=250000