Question

in an ap if the 12th term is -13 and the sum of its first four terms is 24.find the sum of its first 10 terms​

Answer 1

Answer:

Step-by-step explanation:

a12 = a +11d =-13

sum of 4 = (4/2)(a+a+3d)

4a + 6d = 24

2a + 3d = 12

2a + 22d = -26

19d = -38

d=-2

2a -6 = 12

a= 9

sum of 10 terms = (10/2)(a + a +9d)

= 5 * (2a + 9d)

= 5 * (18 - 18)

= 0

Answer 2

☞ In an AP, if the 12th term is - 13.

$a_{n}$ = a + (n - 1)d

Here

$a_{12}$ = - 13

n = 12

$a_{12}$ = a + (12 - 1)d

=> - 13 = a + 11d

=> a + 11d = - 13

=> a = - 13 - 11d ________(eq 1)

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☞ Sum of first four term is 24.

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

Here

$S_{4}$ = 24

n = 4

$S_{4}$ = $\dfrac{4}{2}$ [2a + (4 - 1)d]

=> 24 = 2 (2a + 3d)

=> 12 = 2a + 3d

=> 2a + 3d = 12

=> 2(- 13 - 11d) + 3d = 12

=> - 26 - 22d + 3d = 12

=> - 19d = 12 + 26

=> - 19d = 38

=> d = -2

Put value of d in (eq 1)

a = - 13 - 11 (-2)

=> - 13 + 22

=> 9

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☞ We have to find the sum of first 10 term.

$S_{10}$ = $\dfrac{10}{2}$ [2a + (10 - 1)d]

=> 5 (2a + 9d)

Now put the value of a and d in above eq

=> 5 [2(9) + 9(-2)]

=> 5 ( 10 - 10)

=> 5 (0)

=> 0

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The sum of first 10 term is 0.

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