Examples - Arithmetic Progressions Video Lecture - Class 10

hi there now it's quite common to get
questions in arithmetic progressions
where they give you values of two
particular terms in the sequence and
you're asked to find the first term and
the common difference so in this example
what I want to do is show you how we go
about working these kind of questions
out and what we've got here then is in
an arithmetic progression the sixth term
is two and the tenth term is minus
fourteen and we've got to find them the
first term and the common difference so
to do questions like this we should be
familiar then with what an arithmetic
progression is or an arithmetic sequence
we've got the first term is a and the
second term is a plus one D or just a
plus D third term a + - d 4th term a
plus three D and so on
so the 10th term would be a plus nine D
the fifteenth term would be a plus 14 D
so in general the nth term often called
U n is equal to a plus n minus 1 times D
a formula that you should be familiar
with so we can apply this formula here
then in this kind of question we're told
that the sixth term is two so we're
basically saying then that we're given
that the sixth term or you could write u
6 is equal to 2 and we can substitute
this into this formula here so therefore
what we have got is that - 4 u 6 is
equal to the first term we don't know
that so it's a plus n minus 1 times D
well any 6 so if we subtract 1 that's
going to be 5 and we multiply it by the
common difference D now we've got 2
unknowns in this equation so it's
impossible to solve it so it's going to
require
simultaneous equation so I'm going to
number that equation one so where do we
get our other equation form well it's
clearly from this statement here that
the 10th term is minus 14 so we'll just
put down here then that the 10th term
which is U 10 is equal to minus 14 and
so therefore if we use this equation
again we've got minus 14 equals the
first term a plus n minus 1 n is 10 so
take one away from that gives us line
and times it with D and so that's our
second equation and so it's just a
question of solving these simultaneous
equations and there's many ways that we
can solve these you could do it by
substitution I'm going to do it by
elimination because we've got a just
simply here which can be eliminated if
we do equation 1 minus equation 2
equation 1 minus equation 2 what does
that give us well we've got 2 minus
minus 14 which is going to be 16 and
that's going to equal well the two A's
cancel and then we've got five D minus
nine D which is going to be minus 4 D
and if we divide both sides by minus 4
we end up with D equaling 16 divided by
minus 4 which is that D at common
difference equals minus 4 okay we found
the common difference first rather than
say the first term that doesn't matter
we just go on and find out what the
first term is and we can do that quite
easily by just substituting our value
for D D equaling minus 4 into either
equation 1 or 2 I'm just going to
substitute it into one and if we do that
then we've therefore got that 2 equals a
plus 5 times the common difference which
we now know is my
this fall and so therefore we've got to
equals a minus 20 and if we add 20 to
both sides we end up with that first
term a equaling 2 plus 20 which is 22
now it might be that a question now ask
you to find say a particular term if it
does say at last you'd finds the
fifteenth term you're well equipped to
do that now you could say that the
fifteenth term u15 is equal to a which
will be the 22 plus 14 then times the
minus four okay well I hope it's giving
you some idea then on how we can go
about working out the first term and the
common difference then when we're given
two particular terms in the arithmetic
progression
you