The house on one side of a road are numbered using consecutive positiv

'a' is not necessarily the last house, as there are at least 6 houses.
Also, 'a' must be positive even integer, but in your case a=33.33. Hence your assumption that 6th house is the last house is wrong.

Hello Chetan sir.

Whats wrong with my approach.

I assumed 'a' as the last term .So we can have
a
a-2
a-4
a-6
a-8
a-10

so 6a-30 = 170

6a = 200

a= 33.33[/quote][/quote]


Thanks Nick for replying back.

I agree 'a' is not a number here .The fact that its asking for a range of values .

So,I am looking for starting point in the range for options by considering only 6 houses first.( 'AT LEAST 6 houses.')

IMO approach is not wrong. This approach can be time consuming if you don't get your answer in first 5-6 values of 'n'. You are not sure that for how many values of 'n' you gotta check to reach to the answer.

You can modify your approach a bit though to make it efficient.

Let first house number is 'x'.

n/2[2x+(n-1)*2]=170
n[x+n-1]=170
n is a factor of 170 and is greater than 6.
Possible values of n are 10,17,35 85 and 170.

x is +ve only when n=10











[/quote]


Thanks Nick for replying back.

I agree 'a' is not a number here .The fact that its asking for a range of values .

So,I am looking for starting point in the range for options by considering only 6 houses first.( 'AT LEAST 6 houses.')[/quote]

[quote="nick1816"]IMO approach is not wrong. This approach can be time consuming if you don't get your answer in first 5-6 values of 'n'. You are not sure that for how many values of 'n' you gotta check to reach to the answer.

You can modify your approach a bit though to make it efficient.

Let first house number is 'x'.

n/2[2x+(n-1)*2]=170
n[x+n-1]=170
n is a factor of 170 and is greater than 6.
Possible values of n are 10,17,35 85 and 170.

x is +ve only when n=10


Thanks Nick !!I guess I know where I went wrong now and the fact is same.Taking n as 6 and assuming corresponding numbers is wrong
since both are variable here.

I tried the below approach and I still don't know where am I going wrong .

We can imagine first term as x and so last term will be a=x+(n-1)d =

Now,

n/2( 2x + (n-1)2) = 170

n( 2x + 2n -2 ) = 170

n(x+n -1) = 170

Given that n has to be greater than 6 and n*integer = 170 so n can be 5,10,15.. (5 is ruled out) .For 6,7,8,9 ruled out since 170 has factors(5,2,17)

Now replacing n as 10

10(x+9) = 170

x= 8

Therefore a(last term )= 8 + 9*2= 26

If C is the answer, then 14 and 16 can also be a right ? But if we take 14 or 16 as a we won't get sum as 170. Can you please help me understand how "C" is the answer here ?

No, we have to answer whether a lies in that range, and 16 is in the range given in option C.

Something like if I were to ask you a<0, then what must be true..
a<2 will be true because it gives me that range a<0, but not a<-2.

What you have got is the MAX possible value of a, be it integer or non-integer.
But we are looking for INTEGER value.

I solved following way
given that houses are numbered in consecutive even no
so series ; first be house x then x+2 , x+4 ... so on
for 10 houses we will have ; 10a+90=170 and a= 8
so first house is 8=x
in that case 6th house will be x+10 ; 8+10 ; 18
IMO C

Posted from my mobile device

i tried this way

Given sum=170
Sn=n/2(a1+an);--(1)
Prime factorizing 170=17*5*2
possible combinations in equation (1) can be sn=17*10
apply n=10-->s10=10/2(a1+a10);
s10=5(a1+a10)=170;
now possible value for(a1+a10) is 34 only. Sum of the first and last term has to be 34 to satisfy the above equation.
possible a1=8, n=10, a10=26;
a6=18
option c) is the answer