Question

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

Answer 1

Arithmetic Progression (AP)

• It's general formula = a, a+ d, a + 2d,....

• In an AP :

a is the first term and d is common difference.

Common difference is is find by using formula : $a_{2}$ - $a_{1}$, $a_{3}$ - $a_{2}$,....

The nth term of an AP is given as : $a_{n}$ = a + (n - 1)d

• The sum of 1st n terms of an AP :

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

_______________________________

☞ 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)

_______________________________

☞ 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

______________________________

☞ 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

_______________________________

The sum of first 10 term is 0.

_____________________[ANSWER]

Answer 2

a12 = a + 11d

-13 = a + 11d __(1)

S4 = 4/2 (2a + 3d)

24 = 2(2a + 3d)

24 = 4a + 6d

12 = 2a + 3d __(2)

From (1) and (2)

a = 9

d = -2

S10 = 10/2 (2a + 9d)

S10 = 5[(2 × 9) + 9 × (-2)]

S10 = 5 (0)

S10 = 0