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Dear Students, Kindly find the binder of Grade 10th Maths. Regards, Team askIITians

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one year ago

```							Corrrect answer is option CJatin chandra 21….....................................................................
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one year ago
```							Dear student kc+ k(k-1) = k(c+k-1) = 170 = 2:5:17k is a factor of 170 , and then the constraint k is greater than 6 Limit it to the fllowing numbers10, 17 , 34 , 85, 170 17*16 = 272 which is greater than 170 We must have k = 10 we get c =8 c+10 = 18 Ans is option C ,
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one year ago
```							It's Option cKRUTARTH THAKKARGRADE 10 BINDER MATH.... .    . . . .. . . . . . . . . .. . . . .. . . . . . . .. . . . .
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one year ago
```							Sum of first n positive even numbers is 2+4+6..2n=2(1+2+3+...n)=n(n+1)Clearly n(n+1) cannot be 170, so we can subtract initial numbers one by one, that is adding the numbers one by one to 170..First, 170+2=172=2*86=4*43..not in form of n(n+1).Next, 170+2+4=176=2*88=8*11..not in form of n(n+1)Next, 170+2+4+6=182=2*91=2*7*13=13*14.... in form of n(n+1) so first term is 8 and 6th term is 8+2*(6-1)=8+10=18 CSum of first n positive even numbers is 2+4+6..2n=2(1+2+3+...n)=n(n+1)Clearly n(n+1) cannot be 170, so we can subtract initial numbers one by one, that is adding the numbers one by one to 170..First, 170+2=172=2*86=4*43..not in form of n(n+1).Next, 170+2+4=176=2*88=8*11..not in form of n(n+1)Next, 170+2+4+6=182=2*91=2*7*13=13*14.... in form of n(n+1) so first term is 8 and 6th term is 8+2*(6-1)=8+10=18 Option C  Jatin chandra 21
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one year ago
```							Sum of first n positive even numbers is 2+4+6..2n=2(1+2+3+...n)=n(n+1)Clearly n(n+1) cannot be 170, so we can subtract initial numbers one by one, that is adding the numbers one by one to 170..First, 170+2=172=2*86=4*43..not in form of n(n+1).Next, 170+2+4=176=2*88=8*11..not in form of n(n+1)Next, 170+2+4+6=182=2*91=2*7*13=13*14.... in form of n(n+1) so first term is 8 and 6th term is 8+2*(6-1)=8+10=18 CSum of first n positive even numbers is 2+4+6..2n=2(1+2+3+...n)=n(n+1)Clearly n(n+1) cannot be 170, so we can subtract initial numbers one by one, that is adding the numbers one by one to 170..First, 170+2=172=2*86=4*43..not in form of n(n+1).Next, 170+2+4=176=2*88=8*11..not in form of n(n+1)Next, 170+2+4+6=182=2*91=2*7*13=13*14.... in form of n(n+1) so first term is 8 and 6th term is 8+2*(6-1)=8+10=18 Option CBy Prakhar Gupta Chandra 21
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one year ago
```							Correct answer is option C. ….................................................................................
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one year ago
```							Dhruvi ShahChandra 20Answer is option ‘C’Reason is – If the number of houses are k then let the number on each house be n,n+2,n+4.......n+2(k-1) Sum = 170 = k/2[n+n+2(k-1)] = k/2 (2n+2k-2) = k(n+k-1) Let k(n+k-1) be equation 1Now k is a factor of 170k is greater than equal to 6 (since there are at least 6 houses in the row)now k can be either 10 or 17 or 34 or 85 or 170 (since k is factor of 170)k can not be 17 anything more than that since by substituting it in eq 1 we get the number more than 170So, k=10Sub k = 10 in eq 1,n = 8Number on house 6 = n+2(k-1) = 8+2(5) = 8+10 = 18 So option c is correct
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one year ago
```							a is the 6th number, so a and b option arent valid since the the first house numbers will be negative which is not possible.….that means, a has to be atleast 2 + 2*5 = 12.....now option c means that the first house number will be either 4 or 6 or 8 …. which is possible(since numbers are positive)...hence we check these as to get a sum of 170.. we find that when a is 18, first term is 8.... now suppose there are n terms...so 8+....+{8 + 2*(n-1)} = 170.... this means {2(8) + 2(n-1)}*n / 2 = 170 ....which means that (n+7)*n=170=10*17 → n=10 by subtituting value of n we get first term is 8 and hence 6th number is 18 → a=18 there fore correct answer is option (C)
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one year ago
```							Answer is option c 14By utkal Chandra 22.          Take 6th house as n, then 1st house  as n-10,n-8,n-6,n-4,n-2,n,n+2.....       ....
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one year ago
```							 The sum of all houses is 170 so Sn=170 let the no.of houses be n and x be the no. of first house since numbers are increasing consecutively by even no.s ,d will be 2Sn=n/2[2a+(n-1)d]170=n/2[2x+(n-1)2]170=n[x+n-1]over here n is the factor of 170  170=5x2x17we know there are at least 6 houses so the possibility of n are 10,34,85,17,1700+2+4+6+8.......2(n-1)=n(n-1)10(10-1)=90Hence n is equal to 10 sub n=10 in 170=n(x+n-1)x=8a=8+5x2=18so option c is correct
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one year ago
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