Question

the A.P. in which 4th term is -15 and 9th term is -30. Find the sum of first 10 numbers.

Answer 1

Answer:

-195

Step-by-step explanation:

Given :

In an A.P , 4th term = 15, 9th term = -30 ,.

To find :

The value of Sum of first 10 terms,.

Solution :

We know that,

$a_n = a + (n - 1)d$

&

$a_4 = -15$ and $a_9 = -30$

Hence,

$a_4 = -15$

⇒ $a + (4 - 1)d = -15$

⇒ $a + 3d = -15$ ...(i)

$a_9 = a + (9-1)d = -30$

⇒ $a + 8d = -30$ ..(ii)

By subtracting (i) from (ii)

We get,

$(a + 8d) - (a + 3d) = -30 - (-15)$

⇒ $5d = -30 + 15 = -15$

⇒ $d = -\frac{15}{5} = -3$

By substituting value of d in (i),

We get,

⇒ $a +3(-3) = -15$

⇒ $a - 9 = -15$

⇒ $a = -15+9=-6$

⇒ $a = -6$

__

Hence,

$a_{10}= a + (10 - 1)d = -6 + 9(-3)$

⇒ $-6 - 27 = -33$

∴ $a_{10} = -33$

So, We know that,

⇒ $S_n = \frac{n}{2} (a + a_n)$

⇒ $S_{10} = 5(-6 + (-33))$

⇒ $S_{10} = 5(-6 - 33)$

⇒ $S_{10} = 5(-39) = -195$