Question

If the first term of an AP is 2 and the sum of the first five terms is equal to 1/4 th of the sum of the next five terms ,then find the 20th term?

Answer 1

a=2
S5=5/2(2*2+4d)
S10=10/2(2*2+9d)
given
S5=1/4(S10-S5)
5/2(4+4d)=5(4+9d)/4
4(1+d)=2(4+9d)/4
2+2d=4+9d/4
7d=-2/4
d=-1/14
,A20=2+19(-1/14)
=2-19/14
=9/14

Answer 2

Answer:$frac{-53}{2}$

Step-by-step explanation:

Given that

A=2

As we know that

Sum of n terms = $\frac{n}{2}\left \{ 2a+d\left [ n-1 \right ] \right \}$

and

4*sum of first five terms= sum of next five terms

$</p><p>4*\frac{5}{2}\left \{ 2*2+d*(5-1) \right \}=\frac{5}{2}\left \{ 2*(2+4d)+(n-1)*d \right \}\\\\2*5\left \{4+d*4 \right \}=\frac{5}{2}\left \{ 4+4d+(5-1)*d \right \}\\\\10\left \{ 4+4d \right \}=\frac{5}{2}\left \{ 4+8d \right \}\\\\16+16d=4+8d\\\\8d=-12\\\\d=\frac{-3}{2}</p><p>$so the value of 20th term = a+19d

=$2+19*\frac{-3}{2}\\\\2-\frac{57}{2}\\\\\frac{4-57}{2}\\\\\frac{-53}{2}$