Question

Find the nth term of the A.P
1000, 900, 800 ……………………

Answer 1

Hint: In this question use the direct formula for the nth term of arithmetic progression given by ${a_n} = {a_1} + \left( {n - 1} \right)d$ here a is the first term of the progression, n is the total number of terms and ${a_1}$is the first term of A.P.

Complete step-by-step answer:
The given series of A.P is
1000, 900, 800…
Now we have to find the nth of the A.P
As we know that the nth of an A.P is given as
${a_n} = {a_1} + \left( {n - 1} \right)d$……………………… (1)
Where ${a_n} = $ nth term of the A.P
             ${a_1} = $ First term of the A.P
              d = common difference of the A.P
              n = number of terms.
So in the given A.P,
First term (${a_1}$) is 1000 and the common difference (d) = (900 – 1000) = (800 – 900) = -100.
Not substitute these values in equation (1) we have,
$ \Rightarrow {a_n} = 1000 + \left( {n - 1} \right)\left( { - 100} \right)$
$ \Rightarrow {a_n} = 1000 - 100n + 100$
$ \Rightarrow {a_n} = 1100 - 100n = 100\left( {11 - n} \right)$
So this is the required nth term of the A.P.

Note: A sequence is said to be A.P if and only if the common difference that is the difference between the consecutive terms remains constant throughout the series. It is always advisable to remember all the series related formula whether it is of nth term or of sum of nth as it helps saving a lot of time.