Question

In an AP the 1st term is 2 and the sum of the 1st 5 terms is one-fourth of the next 5 terms.Show that 20th term is -112.

Answer 1

Let the AP
be a1, a2, a3....an
First term is a1=2
Sum of the first five terms=\(a1+a2+a3+a4+a5=5\frac{(a1+a5)}{2}\)
do you get this @JOYAL?

Answer 2

@ash2326 yes, can u complete it?

Answer 3

hard to parse:

"the sum of the 1st 5 terms is one-fourth of the next 5 terms"

but that could just be me

Answer 4

Sum of the next five terms=
\[a6+a7+a8+a9+a10+a11=5\frac{(a6+a11)}{2}\]
@joyal now we have the relation
\[\frac{1}{4}\times5\frac{(a1+a5)}{2}= \frac{(a6+a11)}{2}\]
could you plugin the values of a5, a6, and a11 in terms of a1 and the comon difference d

Answer 5

you might have your 1/4 on the wrong side

if a is 1/4 of b; then, a = b/4

Answer 6

oops @amistre64 Sorry !
\[5\frac{(a1+a5)}{2}\ \\ \ \ \ = \frac{1}{4}\times\frac{(a6+a11)}{2}\]

Answer 7

@amistre64 so u r saying that the question is wrong?

Answer 8

im saying the question is hard to read.

but i follow ashes concept, and do not disagree that it is a plausible rendering :)

Answer 9

@amistre64 @ash2326 thanks for ur help..

Answer 10

youre welcome, so far :)

Answer 11

\[4*5*\frac{(2+a_5)}{2}= 5*\frac{(a_6+a_{11})}{2}\]
if we take that and reduce it we get

\[2(2+a_5)=a_6+a_{11}\]

Answer 12

which i cant see working for us at the moment ....

Answer 13

if we rework the AP in terms of another one

say

\[b_1=(2+a_2+a_3+a_4+a_5)\]
\[b_2=(2+a_2+a_3+a_4+a_5+d)/4\]

this might be promising

Answer 14

after some working out, that thought went nowhere

if we focus on the partials

a1 = 2+0d
a2 = 2+1d
a3 = 2+2d
a4 = 2+3d
a5 = 2+4d
----------
10+10d


a6 = 2+5d
a7 = 2+6d
a8 = 2+7d
a9 = 2+8d
a10 = 2+9d
-----------
10+35d

10+10d = (10+35d)/4

40+40d = 10+35d

5d = 50
d = 10

does; 2+19(10) = -112 ?

since it doesnt, it means that we are most likely reading the problem incorrectly

Answer 15

pfft, forgot how to subtract :) 10-40 = -30; -30/5 = -6

2+19(-6) = -112 ?