30+31+32+33+................ upto x numbers > 12250 then minimum value of x will be

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16 Feb 2012 sum

30+31+32+33+................ upto x numbers > 12250
then minimum value of x will be

OPtion

1) 122
2) 123
3) 125
4) 128
5) 130
6) 132
7) 135
8) 134
9) 138
10) none of these

Solution

Formula for sum of x terms of an arithmetic progression is
S = (x/2)[(x-1)*2nd term - (x-3)*1st term]
= (x/2)[(x-1)*31 - (x-3)*30]
= x*(x+59)/2
so according to the given condition x*(x+59)/2 > 12250
so the minimum possible value of x = 130

Correct Option: 5 Best Solution (15)

Aman   12 years AGO

Sum of x terms = (x/2)*(2*30 +(x-1)*1)
when x=129,
sum = 129*188/2 =129*94=12126
for x=130
sum = 65*189=12285

so min value of x should be 130 so that
30+31+32+33+................ upto x numbers > 12250

Like? Yes (4) | No (1)  

Shiv Kumar   12 years AGO

30+31+32+33+................ upto x numbers, means sum of X numbers starting from 30 is grater than 12250.

Formula To find X is
Hxh+(Lxh)

H=highest number
L=lowest number
h=half of the numbers u want to add
(in this case series starting from 30 and ending with 159 (Total 130 numbers)…divide it in 2, h will become 65)

According to Formula (159*65)+(30*65)= 12285. 12285 is just grater than 12250.

Like? Yes (4) | No (1)  

Alagujeyaraj   12 years AGO

a=30 d=1 n=130
sn=n/2(2a+(n-1)d)
=>65(60+(129)*1)
=>65*189
=>12285

Like? Yes (2) | No (2)  

VAIRAGANESH.V   12 years AGO

if n=128 s(n)=11968;
n=130 s(n)=12285;
so ans= absolutely option 5

Like? Yes (2) | No (2)  

yazar   12 years AGO

a=30 d=1 n=130
sn=n/2(2a+(n-1)d)
=>65(60+(129)*1)
=>65*189
=>12285

Like? Yes (2) | No (2)  

saurabh himanshu   12 years AGO

by using the simple ap series formula by options .use first Sn=n/2(2a+(n-1)d).by taking each of the values from option.

Like? Yes (3) | No (3)  

DIAMOND   12 years AGO

if n=128 s(n)=11968;
n=130 s(n)=12285;
so ans=5

Like? Yes (1) | No (2)  

Anbu.k   12 years AGO

(n)=128
s(n)=11968

then
n(130)
s(n)=12285
so ans is 5 is correct

Like? Yes (1) | No (2)  

jai    12 years AGO

(n)=128
s(n)=11968
then
n=130
s(n)=12285
so ans is 5

Like? Yes (1) | No (2)  

Anbu   12 years AGO

s(n)=11968

then
n(130)
s(n)=12285
so ans is 5 is correct

Like? Yes (1) | No (2)  

Rathika   12 years AGO

s(n)=11968

then
n(130)
s(n)=12285
so ans is 5 is correct

Like? Yes (1) | No (2)  

Pratik Goyal   12 years AGO

reason is the given series is an A.P with common difference 1
and we have to find the last term of A.P
then according to formula of sum of terms of A.P and equals to 12250
and get minimum we have to sum up to 130

Like? Yes (1) | No (2)  

Mullaji Zubair Dawood   12 years AGO

for Summation to be >12250,The no. Upto which is to be Added is from 30 to 159 Hence 130 Numbers.. So Option 5.!

Like? Yes (1) | No (2)  

Matt Damon   12 years AGO

Arithmetic Progression(AP)=n/2(2a+(n-1)d)
a=30;d=1;n=x;
130/2(2(30)+(130-1)(1))>12250
12285>12250

Like? Yes (1) | No (2)  

B.Senthil Kumar   12 years AGO

if n=128, s(n)=12165
if n=130, s(n)12285

Like? Yes (1) | No (4)