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 1

Step-by-step explanation:

given 12th term is -13

12th term = a+ (12-1)d

-13 = a+ 11d

given sum of 4 terms is 24

(4/2)(2a+(4-1)d) = 24

2( 2a+3d) = 24

2a+3d = 12

solving the two equations we get

d = -2 a = 9

now we need to find the sum of first ten terms

sum of 10 terms = (n/2)(2a+(n-2)d)

= (10/2)(2×9 + (8)(-2)) = 5( 18 -16 ) = 5(2) = 10

sum of first ten terms is 10

Answer 2

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

12th term = a₁₂ = -13

→ a + (12 -1)d  = -13

→ a + 11d = -13

→ a = -13 - 11d

Now,

Sum of the four terms = 24

Sₙ = $\frac{n}{2}[2a+(n-1)d]$

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

$\implies 24=12[2a+(23)d]$

$\implies \frac{24}{12}=[2a + 23d]$

Putting the value of 'a' we get

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

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

$\implies 2 = -26 + 1d$

$\implies 2 +26 = 1d$

$\implies 28 = d$

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

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

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

$S_{10}= 5[-642 +(9)(28)]$

$S_{10}= 5[-642 +252]$

$S_{10}= 5[-390]$

$S_{10}= -1950$

Hence,

The sum of the first ten terms is -1950