Question

series
1 If the sum of the first 15 terms of an A.P. be 30 and the sum of first 30 terms is 15,
then find the sum of its first 45 terms.
Hint: Follow the solution of Illustrative Example 11 (b) given on page 408.​

Answer 1

Solution:-

Given :-

$\bf \: \to \: S_n \: = \frac{n}{2} \{2a + (n - 1)d \}$

$\rm \to \: S{_1}_5 \: = 30$

We get

$\rm \: 30 = \frac{15}{2} \{2a + (15 - 1)d \}$

$\rm \: 60 = 15 \{2a \: + 14d \}$

$\rm \: \frac{60}{15} = 2a + 14d$

$\rm \: 2a + 14d = 4 \: \: \: \: \: \: \: \: \: \: \: \: ....(i)eq$

Now take

$\rm \to \: S{_3}_0 \: = 15$

$\rm \: 15 = \frac{30}{2} \{2a + (30 - 1)d \}$

$\rm \: 15 = 15 \{2a + 29d \}$

$\rm \: \frac{15}{15} = 2a + 29d$

$\rm \: 2a \: + 29d \: = 1 \: \: \: \: \: \: \: .....(ii)eq$

Subtract ( i ) - ( ii )

$\rm \: 2a + 14d - (2a + 29d) = 4 - 1$

$\rm \: 2a \: + 14d - 2a - 29d = 3$

$\rm \: - 15d = 3$

$\rm \: d = \frac{ - 1}{5}$

Now put the value of d on ( i ) eq , we get

$\rm \: 2a \: + 14 \times \frac{ - 1}{5} = 4$

$\rm \: 2a \: - \frac{14}{5} = 4$

$\rm \: 2a = 4 + \frac{14}{5}$

$\rm \: 2a \: = \frac{20 + 14}{5}$

$\rm \: 2a = \frac{34}{5}$

$\rm \: a = \frac{17}{5}$

So , we get

$\rm \: a = \frac{17}{5} \: \: \: and \: \: d \: = \: \frac{ - 1}{5}$

Answer 2

Answer:

Step-by-step explanation: