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The sum of n terms of the series 1+1.5+2+2.5+3+...1 + 1.5 + 2 + 2.5 + 3 + ... is?

Solution

✅ Correct Option: 1

The first few terms of the series are:

  • 1st term = 1
  • 2nd term = 1.5
  • 3rd term = 2
  • 4th term = 2.5
  • 5th term = 3

The difference between consecutive terms:

  • 1.5 - 1 = 0.5
  • 2 - 1.5 = 0.5
  • 2.5 - 2 = 0.5

Since the difference is constant (always 0.5), this is an Arithmetic Progression (AP).


For this AP:

First term (a) = 1

Common difference (d) = 0.5


The formula for sum of n terms of an AP is:

Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d]

Substituting the values:

Sn=n2[2(1)+(n−1)(0.5)]S_n = \frac{n}{2}[2(1) + (n-1)(0.5)]

Sn=n2[2+0.5n−0.5]S_n = \frac{n}{2}[2 + 0.5n - 0.5]

Sn=n2[1.5+0.5n]S_n = \frac{n}{2}[1.5 + 0.5n]


Taking out 0.5 as common from the bracket:

Sn=n2×0.5[3+n]S_n = \frac{n}{2} \times 0.5[3 + n]

Sn=n×0.52[3+n]S_n = \frac{n \times 0.5}{2}[3 + n]

Sn=n4[n+3]S_n = \frac{n}{4}[n + 3]

Sn=n(n+3)4S_n = \frac{n(n+3)}{4}

Therefore, the sum of n terms of the series is n(n+3)4\frac{n(n+3)}{4}.

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