Question

The sum of first ten terms of an A.P. = 3 and sum of its first fifteen terms = 16.

Statement (1): The sum of first 5 terms of the given A.P. = 16 − 3 = 13.

Statement (2): The sum of last 5 terms of the given A.P. = Sum of first 15 terms minus sum of first 10 terms.

Options

• Both the statements are true.

• Both the statements are false.

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

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

Answer 1

Statement 1 is false, and statement 2 is true.

Explanation:

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

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

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

⇒ \[S_{10} = 3\]

⇒ \[\frac{10}{2}[2a + (10 - 1)d] = 3\]

⇒ 5[2a + 9d] = 3

⇒ 10a + 45d = 3      ........(1)

Given, the sum of its first 15 term = 16

⇒ \[\frac{15}{2}[2a + (15 - 1)d] = 16\]

⇒ \[\frac{15}{2}[2a + 14d] = 16\]

⇒ \[\frac{15}{2} \times 2[a + 7d] = 16\]

⇒ \[15[a + 7d] = 16\]

⇒ 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 \[{} = S_5\]

\[S_5 = \frac{5}{2}[2a + (5 - 1)d]\]

\[{} = \frac{5}{2}[2a + 4d]\]

\[{} = \frac{5}{2}[2(a + 2d)]\]

= 5[a + 2d]

= 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 \[{} = a_{11} + a_{12} + a_{13} + a_{14} + a_{15}\]

= [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.