Question

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

Answer 1

t12=a+11d=-13;a=first term,d=common difference
sum of 1st 4 terms=a+a+d+a+2d+a+3d=4a+6d=24
=>2a+3d=12
solving we get a=9,d=-2
S10=5(18+9×(-2))
=0

Answer 2

Hi there !

a₁₂ = -13

a + 11 d = -13 ----> (1)

given ,

S₄ = 24

Sn = n/2 [ 2a + [ n-1] d ]
24 = 4/2 [ 2a + 3d ]
24 = 2 [ 2a + 3d ]
2a + 3d = 12 ---> [ 2]

solving equations 1 and 2 ,


(a + 11 d = -13 ) x 2 = 2a + 22d = -26


2a + 22d = -26
- 2a + 3d = 12
============
19d = -38

d = -2

2a - 6 = 12

2a = 18

a = 9

===============================

sum of first 10 terms :-

n = 10
a = 9
d = -2

Sn = n/2 [ 2a + [ n - 1 ] d ]

= 10/2 [ 18 + 9 x -2]

= 5 [ 18 - 18 ]

= 0

sum of first ten terms = 0

============================

first ten terms are :-

we can get the terms by adding -2 to the first term , and then add -2 to the term obtained , and so on .

9 , 7 , 5 , 3 , 1 , -1 , -3 , -5 ,-7,-9

their sum = 0

----------------------------------

Hope this helps you !