1-Find the arithmetic mean of the AP with 41 terms whose 1st term is 2.5 and common difference is 0.75.
2-Find the 15th term of the sequence 20,15,10.........
Hi
Answer to the first question is 17.5 & to the second -50.
Hi
Answer to the first question is 17.5 & to the second -50.
:sneaky:
i have a question of profit and loss can any solve
Hi,
Can someone help me with this problem:
If a, b, c are in A.P., then logx to base a, logx to base b, logx to base c is in ?
1-find the arithmetic mean of the ap with 41 terms whose 1st term is 2.5 and common difference is 0.75.
2-find the 15th term of the sequence 20,15,10.........
17.5 & -50....
1. AP = Sum of terms / Total no. of terms
So, Sum of terms = n/2 = 41/2 = 41*17.5
So, AP = 41*17.5/41 = 17.5
2. Tn = a + (n-1)d
So, T15 = 20 + (15-1)*(-5) = -50
Cheers!
1. AP = Sum of terms / Total no. of terms
So, Sum of terms = n/2 = 41/2 = 41*17.5
So, AP = 41*17.5/41 = 17.5
2. Tn = a + (n-1)d
So, T15 = 20 + (15-1)*(-5) = -50
Cheers!
for the first question 1 can also look it as,
average would be middle number of the AP.
terms are 41, so middle number is 21th
Find 21th term......2.5 +0.75*(21-1) = 17.5....
True!
Can you help me with this:
If a, b, c are in A.P., then logx to base a, logx to base b, logx to base c is in ?
Is Arun Sharma really important... ???
adutt3 SaysIs Arun Sharma really important... ???
Good to solve if u wanna brush up ur concepts. Else practice from anywhr shud be as good.
True!
Can you help me with this:
If a, b, c are in A.P., then logx to base a, logx to base b, logx to base c is in ?
2b=a+c
logb^2=loga+logc
so they are in g.p
wats the oa
1-Find the arithmetic mean of the AP with 41 terms whose 1st term is 2.5 and common difference is 0.75.
2-Find the 15th term of the sequence 20,15,10.........
Hi
Answer to the first question is 12.5 & to the second -50.
1)
A= 2.5
d=.75
n=41
To find arithmetic mean we need= Sum of all the nos/No. of terms
s41= n/2*(2a+(n-1)D)
=41/2(2*2.5+(41-1)0.75)
s41=41/2*35
am = sum/nos
=41/2*32*1/41
=17.5
For 2) We get -50 only if d= -5 ( but why is it?)
i got d= 5
1-Divide 124 into 4 parts which are in AP such that the product of 1st and 4th part is 128 less than the product of the 2nd and 3rd part.
Sol-The basic formulae is same here... but why those 4 numbers cant be a,2a,3a and 4a.
As in book its a-3, a-1, a+1 , a+3
2-The 63rd and 6th terms of an AP are -77 and 37. find the 17th term.
Qn no. 6 : LOD 2..
There are three circular garbage cans each of diameter 2m. the three are touching each other extarnally at one point only.find the circumferance of the rope encompassing the three cans.
Ans : 1. 2pi + 6
2. 3pi +4
3. 4pi + 6
4. 6pi + 6
2pi +6 the three rectangles are formed when their sides are added to give a sum of 2+2+2=6 and the curved part that is wrapped on the sides of the cans upon adding gives the circumference of one circle thus another 2pi.1
Progression
1-Find the common ration of the GP whose 1st and last term are 25 and 1/625 and the sum of the GP is 19531/625.
2-Find the no. of terms in GP whose 1st term is 16 , sum is 1365/64 and common ratio is 1/4.
3-If x
4-If (1^3-t1)+ (2^3-t2)+(3^3-t3)+(4^3-t4)+............(n^3-tn)=n^2(n-3)/4, find tn where t1,t2,t3.....tn are the term of series
1-Find the common ration of the GP whose 1st and last term are 25 and 1/625 and the sum of the GP is 19531/625.
2-Find the no. of terms in GP whose 1st term is 16 , sum is 1365/64 and common ratio is 1/4.
3-If x
4-If (1^3-t1)+ (2^3-t2)+(3^3-t3)+(4^3-t4)+............(n^3-tn)=n^2(n-3)/4, find tn where t1,t2,t3.....tn are the term of series
Could anyone help me out with finding remainder with composite number denominator
Eg. Find remainder 3^32 / 50 or 3^32 / 323
Is their a way we can split 50 into say 25 and 2 or 5 and 10
and a number like 323 can only be split into 17 * 19.
Please if anyone knows how to do these type of questions. 
Puys is there book for arun sharma solution or just the key at the end of the chapters?
abhilas SaysA boy took a seven digit number ending in 9 and raised it to an even power greater than 2000.He then took the number 17 and raised it to power which leaves the remainder 1 when divided by 4.If he now multiplies both the numbers,what will be the unit's digit of the number he so obtains?
Taking all data
**** 9 ^ even power greater than 2000
Multiplied by
17 ^ 5 ( lets assume 5 because it gives 1 remainder wen diviede by 4 )
====
9^2002 x 17^5
( Even power of 9 always end in 1)
( Cyclcity of 7 is 4, so remaineder 1 will give value 7 only)
1 x 7
= > 7 Ans.
Could anyone help me out with finding remainder with composite number denominator
Eg. Find remainder 3^32 / 50 or 3^32 / 323
Is their a way we can split 50 into say 25 and 2 or 5 and 10
and a number like 323 can only be split into 17 * 19.
Please if anyone knows how to do these type of questions.
Do you have the solution with you ??
I tried to solve , I get this answers
with 50 answer = 19
with 323 answer is 270
quidditch88 SaysPuys is there book for arun sharma solution or just the key at the end of the chapters?
Even i am facing this problem.....in the book more than 30% of the question are unanswered , thinking its to easy to be solved.....
But answer key alone wont help!
Could anyone help me out with finding remainder with composite number denominator
Eg. Find remainder 3^32 / 50 or 3^32 / 323
Is their a way we can split 50 into say 25 and 2 or 5 and 10
and a number like 323 can only be split into 17 * 19.
Please if anyone knows how to do these type of questions.
yes
if question like in the format of { A/(B*C) }
where B and C both are coprime
then first A divide by B and get remainder r1 then
A divide by C and get the remainder r2
and then find out least no when divide by A gives remainder r1 and same no divide by B gives remainder r2
for ex
3^32 / 50
first split 50 in to 25 *2
now 3^32 divide by 25 will get remainder 16
and 3^32 divide by 2 will gives remainder 1
now find out the least no when divide by 25 gives remainder 16 and same no divide by 2 gives remainder 1
so ans is 41