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NDA (Held On: 18 Apr 2021) Maths Previous Year paper

Option 2 : -250

Electric charges and coulomb's law (Basic)

41133

10 Questions
10 Marks
10 Mins

**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 = [2a + (n − 1) × d] = (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})

⇒ 4S_{5} + S_{5} = S_{10}

⇒ 5S_{5} = S_{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

⇒ a_{0} = \(\rm-d\over3\)

∴ d = -3a_{0} = -3 × 2 = -6

S_{10} = \(10 \over2\) [a0 + a0 + 9d] = 5[2a_{0} + 9d] = 5[4 - 54], as a_{0} = 2, d = -6

⇒ S10 = 5 × (-50) = -250