207
edits
Changes
Jump to navigation
Jump to search
Line 65:
Line 65: + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Activities - progressions problems (view source)
Revision as of 09:15, 10 July 2014
, 09:15, 10 July 2014no edit summary
#Differnce between first and last terms is given.
#Differnce between first and last terms is given.
#We find the remaining terms of A.P.
#We find the remaining terms of A.P.
==What is asked of student==
To find the remaining terms in A.P.
==Method 1 :- Solution by formula method==
===Assumption===
#He know the general form of A.P.
#He know the simultaniouse linear eq.
==Algorithm==
The general form A.P. is a, a+d, a+2d, a+3d, ......, a+(n-1)d.<br>
But there are only six terms are there so<br>
a, a+d, a+2d, a+3d, a+4d, a+5d.<br>
Sum of six terms is 345.<br>
so a+a+d+a+2d+a+3d+a+4d+a+5d = 345<br>
6a + 15d = 345------------>1<br>
Differnce between last term and first term is 55.<br>
so (a+ 5d) – (a) = 55<br>
5d = 55<br>
d = 11 -------------->2<br>
Now we find the first term of A.P. by substitute in eq 1.<br>
6a + 15d = 345<br>
6a + 15 (11) = 345<br>
6a + 165 = 345<br>
6a = 345 – 165 <br>
6a = 180<br>
a = 30<br>
then, a = 30, d = 11,<br>
so A.P. is 30, 41,52,63,74,85.
===