@albiesriram said:
log x = k(b-c)
log y = k(c-a)
log z = k(a-b)
From the options we see that A, B and C satisfy
@ananyboss said:Q If in any decreasing arithmetic progression, Sum of all its terms except first term equals -36. Sum of all terms except last term equals Zero. Difference between 10th and 6th term equals -16. What will be the first term of this series ???Ans ??????
Diff bet 10th and 6th terms = -16 so common diff -4.
Now (sum of all except first) - (sum of all except last) = last - first = -36 = -4 * 9 so there are ten terms.
Average of 1st 9 terms is 0 so 5th term is 0, so 1st term is 0 - 4*(-4) = 16.
regards
scrabbler
Now (sum of all except first) - (sum of all except last) = last - first = -36 = -4 * 9 so there are ten terms.
Average of 1st 9 terms is 0 so 5th term is 0, so 1st term is 0 - 4*(-4) = 16.
regards
scrabbler
@albiesriram said:
Ans should be option ( C )
a^2 + b^2 = 2[ 1 + cos(a-b) ] = 2 ( 2 cos^2{(a-b)/2} ) = 4 cos^2[(a-b)/2]
1-c/1+c = cos[(a+b)/2]/ cos[(a-b)/2]
= 4 cos [(a+b)/2] * cos [(a-b)/2] / 4 * cos^2[(a-b)/2]
= 2b / (a^2 + b^2)
@pheonix2013 said:is it option c ?? i did by eliminating options
@ScareCrow28 said:Ans should be option ( C )
OA
Hi, @scrabbler
@scrabbler said:Diff bet 10th and 6th terms = -16 so common diff -4Now (sum of all except first) - (sum of all except last) = last - first = -36 = -4 * 9 so there are ten terms.Average of 1st 9 terms is 0 so 5th term is 0, so 1st term is 0 - 4*(-4) = 16.regards@scrabbler said:
Your method is somewhat like arun sharma .... can you explain this step :
Average of 1st 9 terms is 0 so 5th term is 0, so 1st term is 0 - 4*(-4) = 16.
@albiesriram said:
i get the answer as Both C, and D. But by the first equation if a>b then C is probably the answer
@ananyboss said:Hi, @scrabblerYour method is somewhat like arun sharma .... can you explain this step :Average of 1st 9 terms is 0 so 5th term is 0, so 1st term is 0 - 4*(-4) = 16.
I was trying to indicate how it could be done without having to introduce cumbersome variables (i.e. without writing) 😃 
Explanation....theek hai dekho:
Suppose we know that the average of an odd number of terms in AP is x, then the middle term is x. For example if we know that the the total of the first 11 terms in an AP is 143, we can say that the 6th term is 213 without further ado. (Visualise this, don't learn it like a formula. Understand how an AP behaves, else this becomes a difficult problem!)
Hence if the total of 9 terms is 0, their average is 0/9 = 0 and therefore the 5th term is 0.
Now from the earlier info we had already got that common difference d = -4. So 5th term a + 4d = a + 4*(-4) = 0, which means a must be 16.
regards
scrabbler
Explanation....theek hai dekho:Suppose we know that the average of an odd number of terms in AP is x, then the middle term is x. For example if we know that the the total of the first 11 terms in an AP is 143, we can say that the 6th term is 213 without further ado. (Visualise this, don't learn it like a formula. Understand how an AP behaves, else this becomes a difficult problem!)
Hence if the total of 9 terms is 0, their average is 0/9 = 0 and therefore the 5th term is 0.
Now from the earlier info we had already got that common difference d = -4. So 5th term a + 4d = a + 4*(-4) = 0, which means a must be 16.
regards
scrabbler