The sum of first 10 term of an A.P. = 3 and the sum of its first 15 term = 16.

Statement (1): The sum of first five terms of the given AP equals to 16 - 3 = 13.

Statement (2): The sum of last 5 terms of the given AP equals to sum of first 15 term minus sum of first 10 terms.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Let a be the first term of an A.P. and d be the common difference of the A.P.

Using the formula; S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Given, the sum of first 10 term of an A.P. = 3

$\Rightarrow S_{10} = 3 \\[1em] \Rightarrow \dfrac{10}{2}[2a + (10 - 1)d] = 3\\[1em] \Rightarrow 5[2a + 9d] = 3\\[1em] \Rightarrow 10a + 45d = 3 ………………………..(1)$

Given, the sum of its first 15 term = 16

$\Rightarrow \dfrac{15}{2}[2a + (15 - 1)d] = 16\\[1em] \Rightarrow \dfrac{15}{2}[2a + 14d] = 16\\[1em] \Rightarrow \dfrac{15}{2} \times 2[a + 7d] = 16 \\[1em] \Rightarrow 15[a + 7d] = 16\\[1em] \Rightarrow 15a + 105d = 16 ………………………..(2)$

Subtract equation (1) from (2),

⇒ (15a + 105d) - (10a + 45d) = 16 - 3

⇒ 15a + 105d - 10a - 45d = 13

⇒ 5a + 60d = 13.

Sum of first 5 terms = S₅

$S_5 = \dfrac{5}{2}[2a + (5 - 1)d]\\[1em] = \dfrac{5}{2}[2a + 4d]\\[1em] = \dfrac{5}{2}[2(a + 2d)]\\[1em] = 5[a + 2d]\\[1em] = 5a + 10d.$

Since, value of 5a + 10d cannot be equal to 13.

So, statement 1 is false.

The sum of last 5 terms of the given AP = a₁₁ + a₁₂ + a₁₃ + a₁₄ + a₁₅

= [a + (11 - 1)d] + [a + (12 - 1)d] + [a + (13 - 1)d] + [a + (14 - 1)d] + [a + (15 - 1)d]

= (a + 10d) + (a + 11d) + (a + 12d) + (a + 13d) + (a + 14d)

= 5a + 60d.

We have calculated earlier that,

Sum of first 15 term - Sum of first 10 terms = 5a + 60d

Thus, we can say that the sum of last 5 terms of the given AP equals to sum of first 15 term - sum of first 10 terms.

So, statement 2 is true.

Hence, option 4 is the correct option.