Home work answer (10) Series - 1+2+3+4...........+n Bcoz we don't know total number of pages in book so we consider total pages are n This is an A.P series bcoz difference is same i.e 1 We have a= first term =1 Difference, d = 1 Last term,l = n We know sum of A. P series is= n(a+l) /2 So after substitution we get sum is = n(1+n) /2 .... This is the sum of all pages include torn page Its said in questiom that after torn a page the sum of pages become 195 So, total sum of pages including torn page must be greatet than 195 So we chose value of n which gives sum greater than 195 If we choose 10 , sum= n(n+1) /2 = 55 which is less than 195 so rejected Take n = 20 which give sum n(n+1) /2 = 210 which is greater than 195 take n=19 which give sum 190 which is less than 195 so we reject n = 19 So we conclude from this that the total number of pages n=20 bcoz only n=20 gives value greater than 195 and approximate of 195 Now find torn page 210-195 = 15 this means that the torn page sum is 15 Check in option which one give the sum of two pages equal to 15 So only (7, 8 ) gives sum 15 that means the answer is (b) option Correct answer (b) (7, 8) Thank you sir. 🙏
Hw question answer is 7,8 Sum of pages 1-10 = 55 Sum of pages 11-20 = 155 When we add 55+155 we get 210 Then when we subtract the given sum that is 195 from 210 then we get 15 which means page no 7,8 is missing. Thank you sir ❤
@@Shubhamyadav-l1q8t because the sum of the remaining pages is 195, so if we add pages 1-10 = 55 & if we add pages 11-20 = 155, now add both 210, subtract 210-195= 15, Wallah! You got the answer 😅
We've actually followed hit and trial method to check but is not a direct solution...because nowhere in the question it was mentioned about multiples of 10 or such thing.
1+2+.............n (N+1)/2 let put number greater than 195 ten minus 195 let put n=20 20 (20+1)/2 =10*21 =210 minus 210-195 =15 7+8=15 hence option b 7,8
Sir , in Question 4 sum of a1+ a2+ a3+ a15+.....+ a24 is asked which is not in AP as after a3 there is a15 upto a24 (is it a typing mistake?) if not then why did we use Sum of 24 numbers? shoudn't we use sum of 1st 3 terms and then add it to the sum of the numbers from a15 to a24 ? I am confused !!
in this question we can opt another formula i.e. sum= n/2 [ 2a (n-1) d ] in this a= 1 d= 1 and sum is given i.e. 195 put the value and find n 195= n/2 ] 2*1 (n-1) 1] 195= n/2 * (2) (n-1) 195= n* n-1 195+1= n² √196=n n=14 add 1 for that page which was torn so, 14+1=15 (7+8=15) hence opt b is correct
Sir, 21:39 pe, how to judge, ki kaunsi formula kaha use hogi?? Pichhle sums mein sum ki formula alag lagayi gayi thi, aur yaha alag.. yeh pata kaise chalega. Please help
Sir, in Q no -4 Can we do this? 225 = n/2(a+l) here n= 6, as a_1 , 5, 10, 15 , 20 & 24 are 6 terms. 225= 6/2(a+l) (a+l) = 75 Now Sum till a_24 = n/2(a+l) n= 24(atq) Sum= 24/2 (a+l) Put value of (a+l) = 75 Sum= 900
Answer no. 10 ...... Here given that = one page is torn and numbered as usual starting first page as1 and after torn one page no. If page remaining is 195 ...........this all given here information .... From this we can conclude that this became series =1+2+3+4+5.......+n ....here let be no. If total page be n before torn the page .... Now it is clear that it us arithmetic series as difference is 1 so we get get a=1 d=1 and last term =n From this sum of n term =n%2(a+L)= n%2(1+L) =By putting values given we get all this Now we know that after page torn remaining page is 195 so before torn page the no. Should be greater than 195. Now check by putting values Let n=10then 10%2(10+1) =55 which is 55
If we divide 6a+69d= 225 by 3 we get 2a+23d = (225/3) whose value (which is 225/3 ) we put in the formula of sum of a1 to a24 which is equal to 12[2a+23d] so we get 12[225/3] and we solve it simply to get ans
Home work answer (10)
Series - 1+2+3+4...........+n
Bcoz we don't know total number of pages in book so we consider total pages are n
This is an A.P series bcoz difference is same i.e 1
We have a= first term =1
Difference, d = 1
Last term,l = n
We know sum of A. P series is= n(a+l) /2
So after substitution we get sum is = n(1+n) /2 .... This is the sum of all pages include torn page
Its said in questiom that after torn a page the sum of pages become 195
So, total sum of pages including torn page must be greatet than 195
So we chose value of n which gives sum greater than 195
If we choose 10 , sum= n(n+1) /2 = 55 which is less than 195 so rejected
Take n = 20 which give sum
n(n+1) /2 = 210 which is greater than 195
take n=19 which give sum 190 which is less than 195 so we reject n = 19
So we conclude from this that the total number of pages n=20 bcoz only n=20 gives value greater than 195 and approximate of 195
Now find torn page
210-195 = 15
this means that the torn page sum is 15
Check in option which one give the sum of two pages equal to 15
So only (7, 8 ) gives sum 15 that means the answer is (b) option
Correct answer (b) (7, 8)
Thank you sir. 🙏
thanks for solution
thanks yrr
Wonderful 🎉
Thank you✨️
Hw question answer is 7,8
Sum of pages 1-10 = 55
Sum of pages 11-20 = 155
When we add 55+155 we get 210
Then when we subtract the given sum that is 195 from 210 then we get 15 which means page no 7,8 is missing.
Thank you sir ❤
🎉
How you came to know that total n of pages in that books are 20.
@@Shubhamyadav-l1q8t because the sum of the remaining pages is 195, so if we add pages 1-10 = 55 & if we add pages 11-20 = 155, now add both 210, subtract 210-195= 15, Wallah! You got the answer 😅
We've actually followed hit and trial method to check but is not a direct solution...because nowhere in the question it was mentioned about multiples of 10 or such thing.
Clear understanding empowers optimal performance. Together, we'll strive for excellence. Gratitude for your guidance, sir.
1+2+.............n (N+1)/2
let put number greater than 195
ten minus 195
let put n=20 20 (20+1)/2 =10*21 =210
minus 210-195 =15
7+8=15
hence option b 7,8
All concepts are clear. Lectures are very helpful.Thanku sir🙏
Concepts are clear..
Application of concepts is also good
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10th question by use option 195 +x pages
Then takes 1 to suppose 20 pages then sum (1+20)/2 *20 total pages 210
Then 210-195 get our answer
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Sir ,
in Question 4 sum of a1+ a2+ a3+ a15+.....+ a24 is asked which is not in AP as after a3 there is a15 upto a24 (is it a typing mistake?) if not then
why did we use Sum of 24 numbers?
shoudn't we use sum of 1st 3 terms and then add it to the sum of the numbers from a15 to a24 ?
I am confused !!
Same doubt sir.then y 2 a
All clear sir... Thank you so much sir.
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in this question we can opt another formula i.e. sum= n/2 [ 2a (n-1) d ]
in this a= 1 d= 1 and sum is given i.e. 195
put the value and find n
195= n/2 ] 2*1 (n-1) 1]
195= n/2 * (2) (n-1)
195= n* n-1
195+1= n²
√196=n
n=14
add 1 for that page which was torn so,
14+1=15 (7+8=15)
hence opt b is correct
Bhai yrr formula toh check kar lo, aapne 2a ke baad + toh lagaya hi nahi , how it goes correct
Whi to exactly formula
Sn= n/2 [2a+(n-1)d]
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Sir, 21:39 pe, how to judge, ki kaunsi formula kaha use hogi?? Pichhle sums mein sum ki formula alag lagayi gayi thi, aur yaha alag.. yeh pata kaise chalega. Please help
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Thankyou so much for the lecture Sir. HW Ans- 7,8
But how?
sb smjh aagya sirji
thanks for your efforts
Good initiative sir
answer of homework question is( option b) i.e. 7,8
Thanku sir concept clear
Sir, in Q no -4
Can we do this?
225 = n/2(a+l) here n= 6, as a_1 , 5, 10, 15 , 20 & 24 are 6 terms.
225= 6/2(a+l)
(a+l) = 75
Now
Sum till a_24 = n/2(a+l)
n= 24(atq)
Sum= 24/2 (a+l)
Put value of (a+l) = 75
Sum= 900
Very helpful videos
solved sir.
Thankyou
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Homework question
Ye A.P ka "sum" nikalna ka formula aur 'Sum of 1st natural numbers" ka formula same ho gaya.
sir Q4 is something missing I guess
in last it is asking for valuse of a1+a2+a3+a15------+a24
please response.
all clear sir!
Very good sur
Thank you sir 🙏
I have doubt in ap series 4th question we have to find a1 + a2 + a3 + a15....... a23+a25 as after a3 their is a15 how it is in ap series
Answer no. 10 ...... Here given that = one page is torn and numbered as usual starting first page as1 and after torn one page no. If page remaining is 195 ...........this all given here information .... From this we can conclude that this became series =1+2+3+4+5.......+n ....here let be no. If total page be n before torn the page .... Now it is clear that it us arithmetic series as difference is 1 so we get get a=1 d=1 and last term =n
From this sum of n term =n%2(a+L)= n%2(1+L) =By putting values given we get all this
Now we know that after page torn remaining page is 195 so before torn page the no. Should be greater than 195.
Now check by putting values
Let n=10then 10%2(10+1) =55 which is 55
Excellent
All clear❤
Great🎉
Que no 4 mein. a1+a2+a3+a15 ....hai to fir ye AP kese hua, a3 ke bad sidha a15 hai
Exactly
Sir has assumed that it's a1 + a2+......+a24
The answer for the actual question a1+a2+a3+a15+...a24 is 1147.5
Thank you sir...
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Sir, please reconsider question number 4.
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Sir question no.4 is incorrect because you have written a1+a2+a3+a15... ( a15 just after a3 how can it be in AP
Ans. 7&8
sir why n is 135 ...1 is last term na
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Sir ,I want to ask you that why did you divide this equation 6a+69d=225 by 3 in the question no 4 I didn't understand so please clear my doubt.....
If we divide 6a+69d= 225 by 3 we get 2a+23d = (225/3) whose value (which is 225/3 ) we put in the formula of sum of a1 to a24 which is equal to 12[2a+23d] so we get 12[225/3] and we solve it simply to get ans
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Thanks sir
Sir GP me toh infinity ka bta diya aap ne ...agar AP me infinity aya toh...
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