If in AP first term is 1 and the sum of first four term is one-third of of the sum of the next four

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16 Apr 2013 numbers

If in AP first term is 1 and the sum of first four term is one-third of of the sum of the next four term. Common difference of an AP is

OPtion

1) 1
2) 2
3) 3
4) 4
5) 5
6) 6
7) 7
8) 8
9) 9
10) None of these

Solution

given a=1, S4=1/3(S8-S4)
2(2+3d)=1/3((4*(2+7d)-2*(2+3d))
solve
option 2

Correct Option: 2 Best Solution (14)

Pargat Pal Singh   11 years AGO

For first 4 terms of AP,
a=1
d=d,
n=4
last term, l= 1+3d
Sum of n terms = (n/2)*(a+l)
Sum of first 4 terms = (4/2)*(1+1+3d)= 2*(2+3d)
Sum of next 4 terms = (4/2)*(1+4d+1+7d)= 2*(2+11d)
As per given condition
2*(2+11d)= 3*2*(2+3d)
2+11d = 6+9d
2d=4
d=2

Like? Yes (13) | No (2)  

Gunaseelan   11 years AGO

a+a+d+a+2d+a+3d=1/3(a+4d+a+5d+a+6d+a+7d)
4+6d=1/3(4+22d)(Since a=1)
12+18d=4+22d
4d=8
d=2

Like? Yes (10) | No (1)  

Balakrishnan M   11 years AGO

If in AP first term is 1.
Common difference of AP is d.
sum of first four terms = 1+1+d+1+2d+1+3d
= 4+6d
sum of the next four terms = 1+4d+1+5d+1+6d+1+7d
= 4+22d
The sum of first four term is one-third of of the sum of the next four term.
4+6d = 1/3(4+22d)
12+18d = 4+22d
4d = 8
d = 2

Like? Yes (8) | No (2)  

ajitha   11 years AGO

Let us take the ap series
1,1+d,1+2d....1+7d....
the sum of first 4 terms=4+6d
the sum of next four terms=4+22d
since the sum of the first 4 terms=1/3rd of the sum of next four terms
then 4+6d=1/3(4+22d)
d=2
Hence the common difference d=2

Like? Yes (9) | No (3)  

Devendra Marghade   11 years AGO

If a1,a2,....,a8 are the eight terms of AP
If d=difference of an AP, Given a1=1
Sum of first four terms=(n/2)[2*a1+(n-1)d]
=(4/2)[2+3d]= 4+6d ---(i)
The last four terms i.e. 5th,6th ,7th & 8th term of AP will be 1+4d,1+5d,1+6d & 1+7d respectively
So sum of last 4 term=4+22d ---(ii)
Given 4+6d=(1/3)[4+22d] =>4d=8 or d=2

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sanjay kumar   11 years AGO

let first term=a
common difference=d
given a=1
now
given sum of first four term=1/3(sum of next four term)
s4=1/3(s8-s4)
4/2(2*1+(4-1)d)=1/3{8/2(2*1+(8-1)d-4/2(2*1+(4-1)d}
2(2+3d)=1/3{4(2+7d)-2(2+3d)}
4+6d=1/3(8+28d-4-6d)
4+6d=1/3(4+22d)
3(4+6d)=4+22d
12+18d=4+22d
22d-18d=12-4
4d=8
d=2
ans:2

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vigneshwari   11 years AGO

Sum of the first n-terms in A.P = (n/2)[2a+(n-1)d]
Here a = 1.
Sum of first 4 terms = (4/2)[2+3d] = 4+6d....(1)
Sum of next 4 terms = sum of first 8 terms - sum of first 4 terms....(2)
Sum of first 8 terms = (8/2)[2+7d] = 8+28d
From (1) & (2),
Sum of next 4 terms = 8+28d - 4 -6d = 4+22d....(3)
Given, Sum of first 4 terms = (1/3)of sum of next 4 terms
i.e., 4+6d = (1/3)[4+22d]
=> 12+18d - 4 -22d = 0
=> 8-4d = 0
=> d=8/4 = 2
Hence the common difference is 2.

Like? Yes (4) | No (1)  

krishnapriya   11 years AGO

Given a=1
a+a+d+a+2d+a+3d=1/3(a+4d+a+5d+a+6d+a+7d)
(ie)4a+6d=1/3(4a+22d)
Solving we get
4d=8a
put a=1
therefore, d=2

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jatin patware   11 years AGO

terms be 1 1+d 1+2d 1+3d
1+4d 1+5d 1+6d 1+7d
atc
4+6d=1/3(4+22d)
d=2

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Jasaspi Patel   11 years AGO

Hit and trail

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D.Jagruthi   11 years AGO

1st term=a=1
therefore,2nd term =a+d and so on
=>a+a+d+a+2d+a+3d=1/3(a+4d+a+5d+a+6d+a+7d)
=>d=2

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Vinay Sipani    11 years AGO

Sn = (n/2) ( 2a + (n-1)d)

=> 2( 2 + 3d) = (1/3) [ 4(2 + 7d) - 2(2 + 3d) ]
=> d = 2

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sanjay singh   11 years AGO

letc.d.=d and the first termis1
S(4)=4/2(2+3d)=4+6d
S(8)=8/2(2+7d)=8+28d
S(8)-S(4)=the sum of next 4terms=4+22d
A.T.P. 3(4+6d)=4+22d
solving we get d=2
>d=2

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Rats   11 years AGO

4a+6d=1/3(4a+22d)
12a+18d=4a+22d
8a=4d
d=8/4 (since a=1)
d=2

Like? Yes (3) | No (2)