Correct Answer - Option 2 : -250

**Concept:**

Let us consider sequence a_{1}, a_{2}, a_{3} …. a_{n} is an A.P.

Common difference “d”= a_{2} – a_{1} = a_{3} – a_{2} = …. = a_{n} – a_{n – 1}

n^{th} term of the A.P. is given by **a**_{n} = a + (n – 1) d

Sum of the first n terms (S) = \(\frac{n}{2}[2a+(n-1)d)]\)

Also, S = (n/2)(a + l)

Where,

a = First term,

d = Common difference,

n = number of terms,

a_{n} = n^{th} term,

l = Last term

**Calculation:**

Given, a_{0} = 2 ...(1)

S_{5} = \(1\over4\)(S_{10} - S_{5})

⇒ 4S5 = S10 - S5

⇒ 4S_{5} + S_{5} = S_{10}

⇒ 5S_{5} = S_{10}

⇒ **5 × ****\(5\over2\)[2a0 + (n**_{5} - 1)d] = \(10 \over2\) [2a0 + (n_{10} - 1)d]

[∵ n_{5} = 5 & n_{10} = 10]

⇒ 5 × \(5\over2\)[a_{0} + a_{0} + 4d] = \(10 \over2\) [a0 + a0 + 9d]

⇒ 5 × [2a0 + 4d] = 2 × [2a0 + 9d]

⇒ 10a_{0} + 20d = 4a_{0} + 18d

⇒ 6a0 = 2d

⇒ a_{0} = \(\rm-d\over3\)

∴ d = -3a_{0} = - 3 × 2 = - 6

Hence,

S_{10} = \(10 \over2\) [a0 + a0 + 9d]

⇒ 5[2a_{0} + 9d]

⇒ 5[4 - 54]

As a_{0} = 2, d = - 6

⇒** S10 = 5 × (-50) = - 250**