Question

In an AP if the 12th term is -13 and the sum of its four terms is 24 , find the sum of first ten terms ​

Plz give perfect answer I will mark brainliest

Answer 1

Answer:

43.2

Step-by-step explanation:

$12th=a+11d= -13\\or, a=13+11d\\S_{4} = 24= \frac{4}{2} (2a+3d)=2(2(13+11d)+3d)=2(26+22d+3d)=52+50d\\or, 24=52+50d\\or, -28=50d\\or, d= \frac{-28}{50}= -0.56\\or, a+11(-0.56)=13\\or, a=6.84\\or, S_{10}= 5(13.68-5.04)= 5(8.64)=43.2$

Answer 2

QuestioN :

In an AP if the 12th term is -13 and the sum of its four terms is 24 , find the sum of first ten terms ​

ANswer :

The sum of the first ten terms is -1950

Given:

• 12th term of an AP is (-13)

• The sum of the four terms is 24

To find:

• The sum of the first ten terms

SolutioN :

12th term = a₁₂ = -13

→ a + (12 -1)d  = -13

→ a + 11d = -13

→ a = -13 - 11d

Now,

Sum of the four terms = 24

Sₙ $=\frac{n}{2} [2a+(n-1)d]$

$=24=\frac{24}{2} [2a+(24-1)d]\\\\=24=12[2a+(23)d]\\\\=\frac{24}{12} =[2a+23d]$

Putting the value of 'a' we get

$=2=[2(-13-11d)+23d]$

$=2=[-26-22d+23d]$

$=2=-26+1d$

$=2+-26+1d$

$=28=d$

So,

The value of d = 28

Now,

→ a + 11d = -13

Putting the value of 'd' we get

→ a + 11(28) = -13

→ a + 308 = -13

→ a = -13 - 308

→ a = - 321

Now,

Sum of the first 10 terms is

$S_n=\frac{n}{2} [2a+(n-1)d]$

$S_10=\frac{10}{2} [2(-321)+(10-1)(28)]$

$S_1_0=5[-642+(9)(28)]$

$S_1_0=5[-642+252]$

$S_1_0=5[-390]$

$S_1_0=-1950$

Hence,

The sum of the first ten terms is -1950