Question

the first sum of first 10th term of an ap whose 12th term is -13and the sum of the first four terms is 24​

Answer 1

Answer:

0

Step-by-step explanation:

Let first term be a  and common difference be d

given a+11d = -13 ........1

a+a+d+a+2d+a+3d= 24

4a+6d= 24

2a+3d=12............2

solving 1 and 2 we get

a= 9 and d= -2

therefore sum of first 10 terms of the AP

sum = $\frac{n}{2}(2a+(n-1)d$

putting values we

$\frac{10}{2}(2\times9+(10-1)(-2)$

= 0

Answer 2

$S_{10}=0$

Step-by-step explanation:

Given,

12th term of AP is -13

and sum of four term is 24

Let first term is a, and common difference is d,

Solution,

$a_{12}=a+11d$

$a+11d=-13$

$\left ( a+11d=-13 \right )2$       multiply by 2 both sides

$2a+22d=-26$                (1)

Sum of first n terms of an AP is

$S_{n}=n/2\left ( 2a+\left ( n-1 \right )d \right )$

$S_{4}=4/2\left ( 2a+\left ( 4-1 \right )d \right )=2\left ( 2a+3d \right )=24$

$2a+3d=12$                      (2)

    (2)-(1)

$2a+3d-2a-22d=12+26$

$-19d=38$

$d=-2$

$2a+3d=12$            (2)

$2a=12-3d$

$2a=12-3\times -2$            put the value of d=-2

$2a=18$

$a=9$

Sum of first 10 terms of an AP

$S_{10}=10/2\left ( 2\times 9+9\times -2 \right )$

$=5\left ( 18-18 \right )$

$S_{10}=0$