Question

in an ap whose first term is 2 the sum of first 5 terms is one fourth the sum of the next 5 terms show that t20=- 112 also find S20

Answer 1

So Your A.P series is like a, a+d, a+2d,....
where a = 2
In A P 
$S_{n} = \frac{n(2a+(n-1)d)}{2}$
$S_{5} = \frac{5(4+4d)}{2} = 5(2+2d)$
The sum of next 5 terms can be got using
$S_{10} - S_{5} = 5(4+9d) - 5(2+2d) = 20 + 45d - 10 - 10d = 10+35d$
so
$5(2+2d) = 10 + 10d = \frac{10+35d}{4} \\ 40+40d = 10 + 35d \\ 5d = -30 \\ d = -6$
Now
[tex]t_{n} = a + (n-1)d \\ t_{20} = 2 + (20-1)(-6) = -112 \\ S_{20} = 10(4+19(-6)) = 10( 4 - 114) = -1100[/tex]


Method 2:
           So we have a = 2
Sum of first 5 terms = (Sum of next 5 terms) / 4
[tex](a) + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = \\ ((a+ 5d) + (a + 6d) + (a + 7d) + (a + 8d) + (a + 9d) ) \frac{1}{4} \\ 4(5)a + 4(1+2+3+4)d = 5a + (5+6+7+8+9)d \\ 20a + 40d = 5a + 35d[/tex]
we have a = 2
$40 + 40d = 10 + 35d \\ 5d = -30 \\ d = - 6$
Now you can calculate
$t_{20} = -112 \\ S_{20} = - 1100$

Answer 2

Answer:

a20=-112

Step-by-step explanation:

a=2

let common difference be d

a+(a+2d)+(a+3d)+(a+4d)=1/4(a+5d)+(a+6d)+(a+7d)+(a+8d)+(a+9d)

5a+10d=1/4(5a+35d)

(5a+10d)4=(5a+35d)

20a+40d=5a+35d

20a-5a+40d-35d=0

15a+5d=0

divide 3 on both sides

3a+d=0

d=-3a

d=-3(2)

d=-6

the 20th term = a+(n-1)d

= 2+(20-1)-6

= 2+(19)-6

= 2-114

a20= -112