Find the common difference of an A.P. whose first term is 1 and the sum of first four terms is one-t

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12 Jun 2013 AP

Find the common difference of an A.P. whose first term is 1 and the sum of first four terms is one-third of the sum of the next four terms.

OPtion

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

Solution

By given condition a=1 and S4=1/3(S8-S4)
solve (Using Sn=n/2(2a+(n-1)d))
d=2
option 8

Correct Option: 8 Best Solution (16)

Aman   11 years AGO

First 4 terms are 1, 1+d,1+2d,1+3d.

sum = 4+6d
next 4 terms are 1+4d, 1+5d,1+6d,1+7d.
sum = 4+22d
As per condition
4+22d= 3*(4+6d)= 12+18d
4d=8
d=2

Like? Yes (30) | No (4)  

shivendra pratap singh   11 years AGO

according to question
let the first eight terms are a,a+d,a+2d,a+3d,a+4d,a+5d,a+6d,a+7d
(a+a+d+a+2d+a+3d)=1/3(a+4d+a+5d+a+6d+a+7d)
(4a+3d)=1/3(4a+22d)
3(4a+6d)=4a+22d
put a=1
we get
d=2
so option 8)2 is correct answer

Like? Yes (12) | No (4)  

Adarsha Upadhya   11 years AGO

a+(a+d)+(a+2d)+(a+3d)=1/3(a+4d)+(a+5d)+(a+6d)+(a+7d)
4a+6d=1/3(4a+22d)given a=1;
4d=8
d=2

Like? Yes (5) | No (3)  

IFFAT IRAM   11 years AGO

let a be the first term, d be the common difference
sum of first 4 terms=4[2*1+3d]/2 = 4+6d
for the next 4 terms, the first term would be a+4d
sum of the next 4 terms = 4[2*(a+4d)+3d]/2
= 4+22d
according to the problem,
4+6d=(4+22d)/3
12+18d=4+22d
4d=8
d=2
option 8

Like? Yes (3) | No (3)  

lakshmi chaitanya   11 years AGO

optional verification....
1,3,5,7,9,11,13,15
1+3+5+7=16
9+11+13+15=48
16=1/3(48)
hence series= 1,3,5,7,9,11,13,15
d=2

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Ashik Kuppili   11 years AGO

Sum of n terms in ap is n/2(2a+(n-1)d)

Like? Yes (1) | No (1)  

Devendra Marghade   11 years AGO

If a1,a2,a3,.....,a8 are the first eight terms of AP
and d=common difference.
Sum of first four terms=a1+a2+a3+a4
=a1+(a1+d)+(a1+2d)+(a1+3d)=4a1+6d ---(i)
Sum of next four terms=a5+a6+a7+a8
=(a1+4d)+(a1+5d)+(a1+6d)+(a1+7d)=4a1+22d ---(ii)
Given (4a1+6d)=(1/3)(4a1+22d) or 8a1=4d
As a1=1, d=2

Like? Yes (1) | No (2)  

krishnapriya   11 years AGO

given a=1 and sum of first four terms is one-third of the sum of the next four terms
therefore 4a+6d=1/3(4a+22d)
put a=1, we get d=2

Like? Yes (2) | No (3)  

C.Manohar   11 years AGO

1+1+d+1+2d+1+3d=1/3(1+4d+1+5d+1+6d+1+7d)
4+6d=1/3(4+22d)
8=4d
d=2

Like? Yes (1) | No (3)  

dev kumar   11 years AGO

in ap
sum of first 4 term = 4/2[2*1+(4-1)*d]=s(4)
and
sum of all 8 terms = 8/2[2*1+(8-1)*d]=s(8)
so from question given that
1/3*[s(8)-s(4)]=s(4)
by solving this
d=2

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Murali Krishna   11 years AGO

1 3 5 7 9 11 13 15

1+3+5+7=16=1/3(9+11+13+15)

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deep   11 years AGO

s(4)= 4/2[2*1+(4-1)*d]
nd s(8)= 8/2[2*1+(8-1)*d]
so
1/3[s(8)-s(4)]=s(4)
from this eqn.

1/3[4(2+7d)-(4+6d)]= (4+6d)
hence
d=2

Like? Yes (1) | No (3)  

Jalay Shukla   11 years AGO

s(1-4)=1/3{s(1-8)-s(1-4)}

ans formula of s=n/2{2a+(n-1)d}
a=first term i.e 1
n =number of terms 8 or 4 in above equation
d=common difference
put values in formula and ans is 2

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vinod   11 years AGO

(4/2)(2+3d)=(1/3)*2*((1+4d)2+3d))
gives d=2

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Jigyasa Kumar   11 years AGO

Ans. = 2
Shortcut: if the nos. are 1,3,5,7,9,11,13,15
then sum of first four nos is 16 that is 1/3 of the sum of next four nos.
so the common difference is 3-1=2

Like? Yes (3) | No (6)  

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

Like? Yes (1) | No (4)