Question

In ap if the 12th term is -13 and the sum of its first four terms is 24. Find the sum of 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

HEY MATE,
here is the answer
ask if you have any queries
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let the first term and the commen difference be a & d respectively

A/C - I

$a + 11d = -13 \: \: \: \: \: \: \: -(i)$
A/C - II

$(a)+ (a + d ) + (a + 2d ) \\ (a + 3d) \: \: \: \: \: \: = 24$
$=> \: 4a + 6d = 24$

$=> 2a + 3d = 12 \: \: \: \: \: \: - (ii)$

by eq. (i) & (ii)

$= > 19d = - 38 \\ = > d = - 2$
so,
$= > a = 9$
now , sum of first 10 terms

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

$S = \frac{10}{2} (2(9) + (9)( - 2))$

$S = 5(18 - 18)$

$S = 0$
hence,
sum of first 10 terms is 0

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