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`        There are n arithmetiv means between 11 and 53 such that each of them is an integer. How many distinct arithmetic progressions are possible from the above data ?`
11 months ago

1819 Points
```							there must be 7 such aps corresponding to common difference d= 1, 2, 3, 6, 7, 14, 21, which all happen to be factors of 42, as 42= 53 – 11= (n+1)*d
```
11 months ago
SAMBHAV MISHRA
37 Points
```							Answer is 14 bro. .. Try again !
Explain again ? as soon as possible
```
11 months ago
1819 Points
```							my answer is correct. ab tumme hi buddhi kam h to bhai main kya karu. 11, a1, a2, …, an, 53let common diff be d.then 11+d= a1an+d= 53and an= a1+ (n – 1)dor an + d= a1+ ndor 53 = a1+ ndor 53= 11+d+ nd42= d(n+1)d= 42/(n+1)ab maharaj d ek integer h or 42 ko divide karta h, to d sirf 1, 2, 3, 6, 7, 14, 21, 42 hi ho sakta h kuki iske alawa 42 k koi factors hi nahi h. but obviously d cant be 42 as n cant be zero. hence there will be ONLY 7 POSSIBLE VALUES.kuch bhi comment karne or disapprove karne se pehle thodi buddhi ka bhi prayog kar liya karo agar ho tumhare pas. or agar ab bhi tumhe mera answer gakat lagta ho to reason k sath prove krdo or main yahan p tutoring karna chhor dunga.
```
10 months ago
SAMBHAV MISHRA
37 Points
```							first of all mind your language.... and second answer is 14 only.. here is the explanationn arithmetic means are in between a and b; d = b-a/n+1evaluate for how many values of (n+1), d is an integer. @aditya gupta tumhari budhi ghass khane gayi hai !
```
10 months ago
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• 101 Video Lectures
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• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions