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DSSSB TGT Maths Male Subject Concerned- 23 Sep 2018 Shift 1

Option 1 : p^{3}

__ Concept__:

Let a, a + d, a + 2d, …, a + (n – 1) d be an A.P. Then sum of n terms \(S_n=\frac {n}{2}[2a+(n-1)d]\)

Where, a = the first term, d = common difference, n = the number of terms. S_{n} = the sum to n terms of A.P.

**Calculation:**

**Given**:

**S _{n } = n^{2}p**

⇒ n/2 × {2a + (n - 1) d} = n^{2}p

⇒ 2a + (n - 1) d = 2np

⇒ 2a = 2np - (n - 1) d….....(1)

⇒ **S _{m } = m^{2}p**

⇒ m/2 × {2a + (m - 1) d} = m^{2}p

⇒ 2a + (m - 1) d = 2mp

⇒ 2a = 2mp - (m - 1) d…....(2)

From

(1) and (2) , we have

⇒ 2np - (n - 1) d = 2mp - (m - 1) d

⇒ 2p = d (n - 1 + m + 1)

⇒ 2p = d

Substituting d = 2p in equation (1) , we get :

a = p

Sum of a terms of the A.P is given by :

⇒ p/2 × {2a + (p - 1) d}

⇒ p/2 × {2p + (p - 1) 2p}

⇒ **p ^{3}**