- 245
- 525
- 325
- 425
- None of the above/ More than one of the above.

Option 3 : 325

**Given,**

First term of AP = 1

Third term of AP = 15

**Formula:**

**T _{n} = a + (n – 1)d**

**S _{n} = n/2 [2a + (n – 1)d]**

**a = first term**

**d = common term**

**Calculation:**

a = 1

T_{3} = 1 + (3 – 1) d

⇒ 15 = 1 + 2 × d

⇒ 2d = 15 – 1

⇒ 2d = 14

⇒ d = 14/2

⇒ d = 7

Common difference = 14

S_{10} = 10/2 [2 × 1 + (10 – 1) × 7]

⇒ S_{10} = 5 × [2 + 9 × 7]

⇒ S_{10} = 5 × (2 + 63)

⇒ S_{10} = 5 × 65

∴ S_{10} = 325

Hence, option 3 is the correct answer.

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