Question

in an AP whose fisrt term is 2 .the sum of the first five terms is one fourth the sum of the next five terms. show that a20 =-112 and find s20​

Answer 1

Answer:

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

n

=

2

n(2a+(n−1)d)

S_{5} = \frac{5(4+4d)}{2} = 5(2+2d)S

5

=

2

5(4+4d)

=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+35dS

10

−S

5

=5(4+9d)−5(2+2d)=20+45d−10−10d=10+35d

so

\begin{lgathered}5(2+2d) = 10 + 10d = \frac{10+35d}{4} \\ 40+40d = 10 + 35d \\ 5d = -30 \\ d = -6\end{lgathered}

5(2+2d)=10+10d=

4

10+35d

40+40d=10+35d

5d=−30

d=−6

Now

\begin{lgathered}t_{n} = a + (n-1)d \\ t_{20} = 2 + (20-1)(-6) = -112 \\ S_{20} = 10(4+19(-6)) = 10( 4 - 114) = -1100\end{lgathered}

t

n

=a+(n−1)d

t

20

=2+(20−1)(−6)=−112

S

20

=10(4+19(−6))=10(4−114)=−1100

Method 2:

 So we have a = 2

Sum of first 5 terms = (Sum of next 5 terms) / 4

\begin{lgathered}(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\end{lgathered}

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

((a+5d)+(a+6d)+(a+7d)+(a+8d)+(a+9d))

4

1

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

20a+40d=5a+35d

we have a = 2

\begin{lgathered}40 + 40d = 10 + 35d \\ 5d = -30 \\ d = - 6\end{lgathered}

40+40d=10+35d

5d=−30

d=−6

Now you can calculate

\begin{lgathered}t_{20} = -112 \\ S_{20} = - 1100\end{lgathered}

t

20

=−112

S

20

=−1100

Answer 2

Answer:

a5 = 2+4d. --- (1)

s5 = 1/4 (s10 - s5). ----(2)

a20 = a + 19d

-112 = 2 + 19d

d = -114/19

-6

a5 = 2+76

= 78

s20 = 20/2(2-112)

= -1120